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0:00

определение параметра

0:15

пример на понимание что такое параметр

3:35

ДЗ#1

3:55

знакомимся с десмосом

9:20

ДЗ#2

9:37

задача уровня ЕГЭ, решённая с помощью десмоса

19:35

ДЗ#3

19:43

что значит решить уравнение (неравенство) при всех а

24:07

ДЗ#4

24:20

виды условий параметров

29:02

сколько баллов на ЕГЭ может принести параметр

31:35

изучаются ли параметры в школе?

33:07

способ понять условие задачи с параметром (способ с семёркой)

33:24

задача уровня ЕГЭ, решённая с помощью способа с семёркой

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параметр

параметр егэ

егэ

математика

параметр с нуля

параметр с модулем

егэ 2019

егэ 2020

профильная математика

задание 18 егэ математика профиль

профиль

подготовка к егэ

подготовка к егэ по математике

решу егэ

егэ по математике

как сдать егэ

егэ математика

егэ профиль

егэ профильная математика

школа пифагора

пифагор

пифагор егэ

ященко

ященко 2020

5

теорияшколапифагора

00:00:03

behind which the church is hidden

00:00:14

[music]

00:00:16

during the video I will give

00:00:19

homework for what I am talking about there will be 5

00:00:22

dts.

00:00:23

watch the video in full, collect

00:00:25

all the questions from dc and write in

00:00:28

the comments your answers to the test

00:00:29

5 answers should be in ten dots

00:00:33

whoever didn’t put it please put it

00:00:35

and let’s go let’s look at

00:00:38

three elementary linear equations

00:00:40

for example x plus 1 equals 5x plus 3 equals

00:00:49

5 and for

00:00:52

example x plus something else 4

00:00:55

is equal to 5 that is, look in each of

00:00:57

these three equations we

00:00:59

wrote x plus some number is equal to 5 this is

00:01:04

the number the second term it is always

00:01:06

different yes that is 1 first term of

00:01:08

the sum everywhere x and 5 a here is the second number 2,

00:01:11

we add it up everywhere, it’s different,

00:01:13

let’s denote this number with a letter, and that is,

00:01:18

it turns out x + a equals 5 where this is

00:01:23

some number, this is the letter behind which

00:01:25

the number is hidden, depending on

00:01:28

what number we choose, we choose this x and

00:01:30

get it, so let’s let's solve each

00:01:34

equation, that is, x is equal to 5 minus 1, that

00:01:37

is, I think solving this equation here, no

00:01:39

one has any difficulties five minus three

00:01:45

25 minus 41 and how to solve an equation with a

00:01:49

parameter, this is an equation with a parameter,

00:01:51

by the way, if anyone didn’t know,

00:01:52

here it is, how to solve the equation with a

00:01:54

parameter, solving an equation means

00:01:55

simply finding c therefore x is equal to and and the sum of

00:02:00

the deduction they know the term 5 minus and

00:02:04

here is the whole equation with the parameter x

00:02:08

equal to 5 minus and what could be the question

00:02:10

for such an equation, well, for example, like this,

00:02:19

find all the values of the parameter at

00:02:21

each of which the solution to the equation is

00:02:23

negative, look at a is equal to 4

00:02:26

x is equal to 1 prior inside x is equal to 2 when a

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is equal to 1 x is equal to 4 are positive all

00:02:32

the games and what should be a for x to be

00:02:36

negative well, it’s not so difficult

00:02:38

to guess what if a is equal 5x will be

00:02:42

level 0

00:02:43

and if more than 5 it will already be

00:02:45

negative so how to do this and

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we need x to be negative x is

00:02:51

5 minus and so we’ll just describe 5 minus it

00:02:54

should be less than zero

00:02:57

it turns out and more than 5 so with and more than 5

00:03:00

x will be any a 36 substitute

00:03:03

x equals 5 minus 6a 30 on x equals 5 -7 and

00:03:06

negative and so on, solving the

00:03:10

problem with a parameter means finding the

00:03:12

values of a for which x is something, well,

00:03:16

depending on what the conditions require,

00:03:18

that is, solving the problem with a parameter

00:03:20

means to find a parameter, well, super logical,

00:03:22

not everyone understands this and gets confused

00:03:25

because x seems like two unknowns, but I

00:03:27

don’t understand through k, it’s clear, in fact, we

00:03:29

need to find and at which x something is

00:03:32

what is required by the condition, so now to for

00:03:36

this problem,

00:03:37

try to do misinformation number one, here is a

00:03:40

similar problem, there will be five of them

00:03:42

during the video, they will gradually not

00:03:44

appear here is the first dz,

00:03:56

let's look at three graphs of straight lines,

00:03:58

here we have, for example, y is equal to x

00:04:01

plus one increasing straight line with a

00:04:04

slope coefficient of 1 and

00:04:08

intersecting y at the point 1 we have a

00:04:10

game equal to X plus three,

00:04:13

but it is advisable to understand what they look like

00:04:15

and how to construct a graph, that is, this is

00:04:19

grade 7 8,

00:04:21

so let's build all three graphs, you

00:04:24

can do this using a ruler and a

00:04:26

pencil, it will be clear in exams

00:04:28

using a ruler and black gel pens,

00:04:31

but it’s useful to know how to work with the des

00:04:33

mass website, let’s just

00:04:37

build it here for practice, it’s very simple to build here,

00:04:47

well, you can drive in y, and you can not

00:04:50

drive in y, which is equal to just x + 1, here we have a

00:04:53

straight line, then straight line x plus 3 straight line, x +

00:04:59

4 then it is clear that the second term

00:05:04

again we put different ones, yes it turns out that

00:05:07

in general we can say that there is a

00:05:10

straight line x plus a

00:05:11

and on this site you can set any a and

00:05:17

understand how the function will behave,

00:05:18

because it moves if a is equal to 10 then the

00:05:20

straight line will be go through ten there

00:05:23

if a is equal to less and you can

00:05:26

move the graph like this to get some

00:05:30

condition that is required, that

00:05:33

is, so that there is one solution or two

00:05:35

solutions, but in different ways it is

00:05:36

clear that in the exam you will not have a business,

00:05:39

but during training you it’s

00:05:42

convenient and useful to use sometimes so that you just don’t have to

00:05:44

draw every time, and then you’re already in the

00:05:48

exam when you already understand how

00:05:50

graphs move already in the exam in

00:05:53

your head, the business should work

00:05:56

like with trigonometry, here are three

00:05:57

electric circles, someone says,

00:06:00

well, why bother? look at figure 3

00:06:05

genetic crumbs, I won’t have them

00:06:06

on the exam so you look, you train, you understand, and

00:06:09

she’s at cfg,

00:06:10

you’re always the same with graphs, so

00:06:13

let’s leave, for example, only x + a and

00:06:16

x + 4, so we have two straight lines left up to

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x4 we have static a x + 1 usually,

00:06:24

depending on a we get different

00:06:27

different straight lines, what could be the question,

00:06:33

if you have these two straight lines, let’s

00:06:36

try to come up with a question, two straight lines,

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at which and these lines

00:06:44

will, for example, coincide, at which they

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will be parallel, at which and

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intersect, let's answer these

00:06:51

questions, here in general are two straight lines, the

00:06:54

relative position of two straight lines on

00:06:56

a plane, let's remember the reference

00:06:59

material and now we will open the relative

00:07:04

position of two lines on a plane, that

00:07:06

is, if you have k 1x + b 1 1 straight line

00:07:11

and k 2 x 2 second straight line if k are the

00:07:14

same and would be the same, then the straight lines

00:07:17

coincide, for example 2x + 7 and 2 x 8 minutes

00:07:19

from the same thing, it’s clear that I’m slightly

00:07:21

the same. to the center of redex the same

00:07:23

abe different then the lines are parallel well, for

00:07:27

example, in the same business you can

00:07:28

see how it will look if

00:07:30

to different all the lines intersect no

00:07:32

matter what there s.b. that

00:07:34

means if I have two straight lines

00:07:38

for example y equals so well these straight lines

00:07:42

to in general form can be written as y

00:07:43

equals x plus some number and so

00:07:48

if I take only two straight lines x + 4

00:07:50

for example x + 4 and y equals x plus a

00:07:56

and let's formulate a very simple

00:08:01

question,

00:08:05

find all the values of the parameter for

00:08:08

each of which the lines coincide, well,

00:08:12

how to solve this, that is, the lines coincide

00:08:14

when they have k equal and b equals k equals

00:08:17

one and here and there for equality b in

00:08:21

general for equality we could influence

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if a is equal to 4, too, that is, we will have

00:08:25

someone 4 x 4,

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so what is the answer here we write that a is

00:08:33

equal to 4, well, in general, the answer is equal to 4 to

00:08:37

about four straight lines coincide, and if, for

00:08:40

example, such a question, find all

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the values of the parameter for each of which

00:08:47

lines are parallel, how can we have straight lines in

00:08:56

parallel if k are the same abe are

00:08:58

different because the lines were

00:09:00

parallel to we have the same so if

00:09:05

a is equal to 4 then the straight lines will coincide

00:09:08

and I need them to be parallel, that

00:09:09

is, here is the answer what is the answer here and

00:09:11

just not a four anything except four

00:09:14

and the lines will be parallel if a is equal to

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4 then the lines coincide if they are 4 then the

00:09:20

lines are parallel and let’s for this

00:09:23

simple task,

00:09:24

yes for number two we go to find all the values of

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the parameter for each of which the lines

00:09:32

intersect slightly others,

00:09:38

let’s analyze the USE level problem

00:09:41

that came up on the Unified State Examination,

00:09:44

well, it has appeared in the demo version for 6 years

00:09:47

in a row, and in Yashchenko’s books,

00:09:50

it also often appears, this is the problem,

00:09:53

let’s solve it using dice mass, that

00:09:55

is, we have a system of two equations,

00:09:57

we need to find such and for each of which

00:10:00

the system has one solution the first

00:10:04

equation specifies two circles, the second equation

00:10:07

gives the circle a floating radius and

00:10:09

we need to find such a and at which there will be 1

00:10:12

intersection, that’s all, well, the module can

00:10:17

open up as x modules of the registry office or maybe as

00:10:19

minus x, so we get two

00:10:22

circles from the first equation, two

00:10:24

circles are obtained in general the equation

00:10:26

of a circle, if anyone doesn’t remember, let me remind you

00:10:27

that x minus x zero squared plus y

00:10:34

minus y 0 squared is equal to the radius

00:10:37

squared, that is, the first equation

00:10:40

is actually a circle with a center,

00:10:45

let’s say so x minus 5

00:10:48

squared plus y minus 4 squared is equal

00:10:53

3 squared, that is, this circle with

00:10:56

center 54, here they are, the coordinate of the center is 54

00:11:00

and the radius is 3, and if the module language c

00:11:06

opens with a minus, that is, minus x

00:11:10

minus x minus 5 squared,

00:11:12

this square is taken out minus will kill and it turns out x

00:11:15

plus 5 squared plus y minus 4

00:11:22

squared is equal to 3 squared, that is, this

00:11:25

circle with a minus radius with a center

00:11:28

-54 and a radius of 3

00:11:30

and this second is a circle

00:11:37

we have with a center -20 and a radius and can

00:11:44

say the modules for well, but they should be

00:11:45

positive, even they tell us everything

00:11:47

is positive, that is, the radius can simply be

00:11:49

taken like this, let’s build,

00:11:53

let’s build this circle,

00:11:55

this circle, and then a floating

00:11:56

circle as in r mixture, build a

00:11:58

circle like this x minus 5 squared plus

00:12:00

y minus 4 squared by 3 squared x

00:12:07

minus 5 press shift and six get

00:12:12

into the exponent squared like this plus

00:12:16

y minus 4 squared

00:12:18

y minus 4 shift and six equals 9

00:12:25

here we have the first circle further 2

00:12:29

circle 2 circle that is, the first

00:12:35

equation defines the blue and green

00:12:36

circle, well, what's up with the second

00:12:39

equation the second equation starts at

00:12:43

-20 and has a floating radius of -2 so x

00:12:47

plus 2

00:12:48

andrey t plus 2 squared here is just y

00:12:56

and here a squared

00:13:02

and depending on what kind of

00:13:05

circle we have there right now for

00:13:08

example has no solutions with the first

00:13:12

equation now she has one touch

00:13:15

to one solution there is now a solution 2

00:13:18

now a solution 3 now a solution 444

00:13:22

now solutions 3 a

00:13:24

solution 2 2 2 2 2 2 2 2 2

00:13:27

you understand what is happening and a little

00:13:30

further there will be one solution, look at

00:13:31

me and there are more than ten it’s not set out how

00:13:33

to do this, because it’s like this right here on

00:13:37

this button like this like this like they’re not

00:13:40

on this button you need to

00:13:41

press the border with the noses, let’s increase it a little

00:13:44

more, I have 10, let’s

00:13:47

put 20 I don’t know, we have 1 to solve

00:13:55

yet since

00:13:56

if a is equal to zero then the circle with a

00:13:59

floating radius generally turns into

00:14:02

a point, well, naturally the solution will be 0

00:14:06

intersection blue green

00:14:08

0 further solution 000 one solution because

00:14:13

one touch here it is further solution 2 2 2 2 2 2 2 2 2 3 4 4

00:14:18

4 4 4 4 4 4 3 2 2 2

00:14:26

2 2 solutions 2 2 2 2 2 1 0 0 that is, one

00:14:30

solution when we fixed here and

00:14:33

for a is equal to 2

00:14:35

now we will prove why there are two so let’s

00:14:42

then add a

00:14:45

red circle

00:14:48

and for a it is not clear how many

00:14:51

here 11 1 but it’s unlikely to be 11 1 it’s

00:14:55

most likely all some kind of irrational

00:14:56

value, let’s find it, that

00:14:58

is, one solution only in these two

00:14:59

cases will be, in principle, you can

00:15:04

keep this graph for yourself, you can

00:15:07

have a function here, you can make a link

00:15:10

to link and copy and

00:15:12

send somewhere you can export the file Well, in

00:15:14

my opinion there will be some kind of problem with

00:15:17

single segments and here you

00:15:21

can try to solve it somehow, in principle,

00:15:26

in principle, it seems like it will be fine now

00:15:28

let’s try, although some huge

00:15:35

graph turned out wrong I don’t want

00:15:38

let’s know how to do it just the

00:15:42

old fashioned way print screen we

00:15:49

need to find in general for what and there will be

00:15:55

one solution now let’s find here

00:16:01

we have a graph so look the first

00:16:11

equation defines two circles

00:16:12

green and blue

00:16:14

here their centers are the center of the

00:16:21

circle with a floating radius 1 time

00:16:26

one the solution will be when we touch the

00:16:28

green and red circles, the point

00:16:33

of contact will lie on the line

00:16:36

connecting the center of the circle, here it is.

00:16:39

by touching this way we need to find the radius of the

00:16:43

red circle, but we realized that this is

00:16:45

equal to 2, but let's prove why it is 2,

00:16:47

well, because this segment we have

00:16:51

four cells, and this segment

00:16:56

is 3 cells according to Pythagoras, we can find

00:17:04

this segment

00:17:05

along the mountains, it will be 5 but this segment

00:17:09

is equal to three because the radius of the green

00:17:12

circle is equal to three and therefore how much does it

00:17:14

turn out to be, how much is the radius of the small

00:17:15

circle, well, two is clear, that is, the first

00:17:19

answer is equal to 2,

00:17:20

find everything positive for each of

00:17:23

which the system has a unique

00:17:24

solution a is equal to 2, so 1 load

00:17:26

now what is there in purple, well, the same

00:17:28

thing, that is, we have purple blue

00:17:31

touching, the point of contact lies on the line

00:17:38

connecting the center of the circle and we need to

00:17:41

find the radius of

00:17:43

the purple circle, how to find it,

00:17:47

so let’s find the segment here,

00:17:50

first from the triangle a b c

00:17:53

rectangular bc equals 4 cells

00:17:56

father equals 7 cells obep fagor we find

00:18:00

so 4915

00:18:02

root 65 turns out b.d.

00:18:06

a b d is equal to three because the radius of the

00:18:09

blue circle is 3, for example, we

00:18:12

were looking for

00:18:13

a.d. we were looking for the radius of the purple

00:18:15

circle, it turned out to be the root and 65 plus 3

00:18:18

is exactly 11 and a little, so the

00:18:22

root is 65 plus 3 again what question

00:18:25

find everything and for each of which

00:18:27

the system has a unique solution,

00:18:28

we have two, this is the

00:18:29

problem like this, it will

00:18:31

help us solve the problem, that is, without it, how

00:18:34

would we do it, but we could easily build the

00:18:36

blue and green truths,

00:18:38

and without a compass in the game, it wouldn’t be

00:18:41

very elegant in appearance, but nothing, and then

00:18:46

we would move the small

00:18:48

circle, we would increase its radius,

00:18:51

well, two we If we could easily get the

00:18:53

violet circle of radius

00:18:57

using coordinates, we would still find it, but

00:18:59

the drawing would of course not be so

00:19:00

beautiful, but in the d sense we can clearly

00:19:03

see that we haven’t lost a single solution,

00:19:05

that in fact there

00:19:07

will be only one solution in these two cases, we even

00:19:08

looked at when it will be there you create two

00:19:10

solutions 340 in general, all cases

00:19:13

have been considered here is such a problem, let’s

00:19:18

now have a similar one for you, we will have it

00:19:22

already up to number 3

00:19:39

you for number three one of the types of problems with a

00:19:46

parameter that is rarer but nevertheless

00:19:48

it is to solve the equations or inequalities for

00:19:52

each and let’s figure out what this

00:19:55

means using an example, here is such a task

00:19:57

for all and solve the inequality this is not the Unified State

00:20:01

Exam level, it’s clear it’s too easy but

00:20:04

such a reference problem for the parameters how to

00:20:07

solve it, well, look on the right it’s not clear

00:20:10

what’s on the right maybe either a zero

00:20:12

or a negative number or a

00:20:13

positive number on the right some kind of

00:20:15

uncertainty the root is greater than

00:20:17

the uncertainty how to solve this

00:20:19

let's look at different types of

00:20:21

uncertainty that is, if we have

00:20:24

a minus and for example less than zero, that is,

00:20:28

the situation is such that the root of x is greater than

00:20:31

which of the negative number x then

00:20:36

should simply satisfy the dose of

00:20:38

tuzik will be greater than or equal to 0 and then in

00:20:40

any case it will be greater than -2

00:20:42

you understand, that is, there

00:20:44

cannot be a negative number under the root, but

00:20:47

if there is a zero under the root ok, if under the root there is

00:20:49

one ok 2 ok 3 ok and so on, that

00:20:52

is it turns out x is greater than or equal to 0,

00:20:59

if the minus, for example, is equal to zero, then

00:21:02

what happens is the root of x is greater than a

00:21:05

zero, which can be and xyg if x is a

00:21:09

zero, this does not suit us because it

00:21:11

must be strictly greater than zero, if

00:21:13

a one is a two, and so on, then it’s suitable

00:21:15

is there a solution here for x x

00:21:18

it turns out strictly more than me should

00:21:20

be and if the minus is greater than zero, that is, for

00:21:25

example, the root of x is greater than 2

00:21:30

then what should x be?

00:21:37

41 is greater than 2

00:21:40

to these, roughly speaking, we square both sides,

00:21:44

that is, it turns out x is greater than

00:21:50

minus a squared,

00:21:56

either the

00:21:57

first case or 2 or 3, well, let’s

00:22:00

solve this, it turns out if a is greater than 0

00:22:06

then x greater than 0 if a is equal to zero

00:22:11

then x is greater than zero if less than 0x is greater

00:22:18

than a square in principle a is equal to zero

00:22:27

could be included

00:22:30

with a is less than or equal to 0 in principle

00:22:35

let's write it down separately that is, what is the answer for

00:22:36

all a solve the inequality the

00:22:38

answer is obtained for a and let's

00:22:49

from minus infinity

00:22:50

we go to plus infinity when a is less than zero,

00:22:52

we can write when a is less than zero or when a

00:22:55

is from minus infinity to

00:22:58

0, we understand that x we have is greater than a

00:23:06

square x belongs from a square to

00:23:09

plus infinity when

00:23:13

a is equal to 0x

00:23:17

we have from 0 to plus

00:23:21

infinity with and with the proper from 0

00:23:32

to plus infinity x we

00:23:36

have from 0 inclusive to plus

00:23:41

infinity that is, once again what is it to

00:23:43

solve for all and this means for all

00:23:47

existing one

00:23:48

we look here in the union we

00:23:51

considered all numbers in general that is before

00:23:53

0 0 and after 0 this is generally all and all

00:23:55

real numbers so for each a we

00:23:59

solved the inequality for each a we said

00:24:01

what x is equal to this means for all a

00:24:03

solve the inequality it’s clear try

00:24:06

to do a similar problem

00:24:08

dz number 4

00:24:22

conditions for problems with a parameter most often

00:24:25

occur two types of

00:24:26

two types, let's write down the conditions of problems

00:24:31

with a parameter, well, let's say the condition of

00:24:35

task number 18, I would say, of course,

00:24:40

there are completely different problems with a

00:24:42

parameter, especially in some kind of

00:24:43

books or manuals, but formats and g as a

00:24:46

rule 22 there can be all kinds of

00:24:49

conditions in the word task 18 is to find all and

00:24:59

for each of which

00:25:01

or there for which the equation or

00:25:08

inequalities has something so for

00:25:11

which the equation

00:25:12

for example well or there the inequality the system

00:25:15

does not matter has some roots

00:25:23

one or two or three roots or has at

00:25:26

least one this mini two solutions or

00:25:31

has one root on the segment,

00:25:32

you understand, yes, that is, find all for

00:25:35

which the equation has,

00:25:36

well, something there, how many to the roots, relatively

00:25:38

speaking, this is the first first type of condition

00:25:43

for task number 18 and the second type is for

00:25:46

all, and solving the inequality equation is

00:25:49

most often the task of one of the two types

00:25:51

falls to the 18th position for all and solve

00:25:59

equations or inequalities

00:26:09

most often they present this type, today we looked at

00:26:12

both at the beginning, so we

00:26:16

were looking to find all for each of which the

00:26:17

solution is negative, for example, this is the first

00:26:21

type, find all we will order from which

00:26:24

straight lines the first type is essentially the same,

00:26:26

yes, find all for each of which

00:26:28

the system has one solution, the first type

00:26:31

for all, and solve the inequality,

00:26:34

this is the second type, that is, we have analyzed both

00:26:37

once again, let’s establish what a

00:26:42

parameter is, a

00:26:44

parameter is the letter behind which

00:26:47

which one is hidden -that’s a number and now you’re thinking about what

00:26:50

number to take so that

00:26:53

the conditions are met in groups of grief, we

00:26:55

see the conditions of the problem, for example this one, and

00:26:58

we select such a value for the hidden

00:27:01

number behind the letter so that the

00:27:04

condition of the problem is fulfilled so that,

00:27:05

well, here for example, yes, we are looking for something like this and

00:27:08

what happened one solution is the most important thing,

00:27:16

further problems in misunderstanding

00:27:18

the parameters may be associated with your

00:27:20

gaps in the seventh-eighth ninth

00:27:23

tenth and eleventh grade,

00:27:24

that is, if you, for example, do not understand well

00:27:28

about the shifts of graphs, for example, when the

00:27:31

pair was narrow when it was wide, like this in the

00:27:34

equation of a parabola you

00:27:35

can see when it moves up when

00:27:39

to the right down to the left about other graphs, you

00:27:44

also need to understand that when the check mark

00:27:46

is expanded, the language module with a graph about straight

00:27:49

lines, the relative position of straight lines, in

00:27:51

general, some basics in well, for the most

00:27:53

part, this is the eighth grade,

00:27:55

but if with these basics of problems it’s

00:27:57

graphic your method will fall a little

00:27:59

and the analytical method is also

00:28:01

mostly eighth grade, 9th grade,

00:28:03

that is, problems in misunderstanding of the parameter,

00:28:08

these are most likely some gaps

00:28:11

there 8 in ninth grade and

00:28:13

insufficient visibility, that is,

00:28:14

if a person is for the first time in his life opens

00:28:17

the parameter, he may be

00:28:19

scared and not understand anything, and when

00:28:21

you are there 10 parameters, look, you

00:28:24

will already have a chance to solve,

00:28:25

listen to me, please, that is, if you

00:28:30

are training, you have a chance to solve such a

00:28:34

question, I will ask, think about what is the

00:28:37

correct answer, how to learn to solve the

00:28:41

parameters, somehow you think the answer may

00:28:48

seem banal, but

00:28:49

it’s really true;

00:28:51

in order to learn how to solve parameters, you need to

00:28:56

solve them in an exam; they’re all worried

00:29:00

about 2; if you fully understand and

00:29:06

understand about ten parameters, for example,

00:29:08

with the help of my free streams with

00:29:11

analysis of options every Monday,

00:29:12

then you will already have good chances to solve

00:29:15

this problem in the game, maybe not

00:29:18

completely, maybe partially,

00:29:19

but maybe completely, but based on and by

00:29:22

2019, an elementary parameter fell out

00:29:24

and many solved it, but here who has

00:29:27

never solved it, he solved no, he didn’t solve it,

00:29:29

and he didn’t didn’t understand, he didn’t know that

00:29:33

this was an easy parameter, the one who had at least

00:29:35

ten times solved it, he already understands

00:29:38

that it was an easy task, it can be

00:29:42

solved at least for an incomplete point, it was definitely

00:29:44

easy to solve, it was a

00:29:49

very significant task, we can

00:29:51

add to you a lot of points, look

00:29:54

how many points the parameter weighs, the

00:29:57

parameter weighs 4 primary points, what does this

00:29:59

mean, this means that in

00:30:02

certain

00:30:03

scenarios, you can add yourself plus

00:30:09

23 test points, how do you like this

00:30:11

information, for example, how does it work,

00:30:14

for example, you solved everything in the first problem

00:30:17

7 problems in the first part in the first part

00:30:20

solved 7 problems out of 12, well, sometimes

00:30:24

7 problems were solved and solved 13 the

00:30:26

problem is completely

00:30:27

13 hanging two primary ones, that is, you

00:30:30

scored

00:30:31

seven plus two, you scored 9 primary

00:30:35

points, that is, so far your test score is 45, well

00:30:40

45 is not very much, but if you came to the

00:30:45

main wave, for example, like in 1919,

00:30:47

you got an easy parameter and you once

00:30:49

tried to solve it a little, you

00:30:51

saw that it was easy to solve it

00:30:53

and you scored, it turns out to be 13 primary

00:30:57

because it weighs 4 primary you

00:31:00

scored 13 primary t68 test tests instead of

00:31:03

45 you scored 68 how do you like this information

00:31:07

plus 23 of one problem how do you like this how

00:31:12

can you hear what I just showed at least

00:31:14

sometime in your life

00:31:16

skip completely when solving a variant of a problem with a

00:31:18

parameter always spend at least

00:31:21

five minutes to get into it somehow poke around

00:31:23

try something there by laying out

00:31:25

by squaring or building

00:31:27

because it may depend on this whether you

00:31:29

enter in darkness

00:31:30

or in sweat are the parameters studied at

00:31:38

school is a common question that I get asked and

00:31:41

some even say not even a question

00:31:44

they ask complain Evgeniy, can you imagine, we do

00:31:48

n’t study the parameters at school, they have

00:31:49

types, imagine what we should do in

00:31:51

reality, yes, you may not be

00:31:55

solving problems in number 18 from the game for months

00:31:58

because the school curriculum is needed to

00:32:02

prepare you for mathematics, they are for the Unified State Exam,

00:32:06

but if you were preparing good at

00:32:10

mathematics 11 years old, then you know the whole

00:32:13

theory to parameters, that is, the theory to

00:32:15

parameters is 8 9 10 part of the class, that

00:32:21

is, these are shifts of graphs, know everything

00:32:24

about the quadratic trinomial and the Vieta theorem,

00:32:27

well, some of the most elementary basic

00:32:29

things, there is nothing beyond the ordinary in the

00:32:31

school curriculum, nothing like that is clear,

00:32:33

there is a parameter that is more complicated and really

00:32:36

very difficult there at

00:32:37

level 1, but in the game such people usually do

00:32:39

n’t fall into it, especially on the main wave,

00:32:41

but lately it wasn’t

00:32:43

all that straightforward and everything was scary, it was quite

00:32:46

classic, so don’t be scared and don’t

00:32:49

It’s worth giving up on this problem in general, it’s

00:32:51

worth looking at it periodically and

00:32:54

trying to solve it when you train, you

00:32:57

understand them better and better,

00:32:59

you feel these parameters, you understand,

00:33:00

so you just try to solve the problem with the

00:33:03

parameters every time you see it, and in

00:33:06

general there is the most common problem with the problem

00:33:10

with the parameter when you see conditions,

00:33:13

you don’t understand what to actually do,

00:33:15

here’s what’s required: there are so many

00:33:17

unknowns, I don’t understand, like and

00:33:21

if you feel this way, for example, for

00:33:24

example, here’s a task from the main

00:33:27

wave of 16, here’s a person who

00:33:31

hasn’t prepared much for the parameters or is just

00:33:33

starting, he’s watching for this and such x

00:33:37

and here there are two unknowns more than one,

00:33:40

as I’m used to, it’s already difficult, it’s not for

00:33:43

me, I’ll probably skip the task to her and oh

00:33:47

well, let me not have plus 4

00:33:49

primary points that can come in

00:33:51

23 tests, well, that’s okay, but no

00:33:53

mine this is not destiny,

00:33:54

but in fact this is a very simple

00:33:57

equation, if it still doesn’t

00:34:01

seem simple to you, I can simplify it,

00:34:03

there is a method that I came up with with

00:34:06

seven, a method with seven, I remind you that the

00:34:14

parameter is the letter behind which

00:34:16

the number is hidden, that is, and this is the letter behind which

00:34:18

what -then the number is k

00:34:19

but let's take seven, that's

00:34:22

my method, that is, for a while I just substitute the water for

00:34:25

seven, it turns out the root, there

00:34:27

was such a terrible equation will be the

00:34:29

root of 3 x square plus if a is equal to

00:34:32

seven 14 x plus 1 is equal to x square plus

00:34:38

7x plus there is only one unknown

00:34:42

left, don’t you understand

00:34:44

where to start solving such an equation,

00:34:46

does someone really not understand

00:34:49

equation 18,

00:34:53

the problem has become level 5 in principle, problems

00:34:55

and g, that is, the level of the first part of the

00:34:57

elementary problem, how to solve the equation, the

00:35:00

root is equal to something unclear, well, square

00:35:05

both parts but not quite square both

00:35:06

parts, of course, as we solved today

00:35:08

with inequality there in front of everyone, but solve in

00:35:11

general, I think that it has become much

00:35:13

clearer, you must agree, that’s who was scared by the

00:35:16

initial condition, he didn’t understand where to

00:35:17

start, doesn’t this person

00:35:19

understand why start here too, but

00:35:21

maybe he doesn’t understand to start

00:35:23

and here it means he has deep

00:35:24

problems with rational equations there,

00:35:26

this is again 8th grade, for the most part

00:35:30

problems with a parameter are 8th grade, since

00:35:35

solving means we say that the right

00:35:38

side should if the right side

00:35:39

negative, well, straight -2, the root is equal to

00:35:42

minus 2 if there is no such solution, if the

00:35:46

right side is equal to zero, then the radical is

00:35:48

also 0, good if the right side is more than 0

00:35:52

ampere 2, then the root is 4, that is, if the right

00:35:59

side is greater than zero, then the radical must

00:36:01

be greater than this number squared in short,

00:36:03

in principle, I think we

00:36:08

understood how to start by substituting seven,

00:36:10

and let's solve this problem, that

00:36:18

is, we say that the right

00:36:22

side cannot be negative, the root cannot

00:36:24

equal a negative number, that is,

00:36:25

we say that x square plus bx plus

00:36:29

one must be greater than or equal to 0

00:36:31

the task is completed in the main 16 years,

00:36:34

that is, this is the absolute level and r so

00:36:38

if the right side is non-negative then,

00:36:41

roughly speaking, we square both

00:36:43

sides 3x squared plus 2x + 1 equals x

00:36:51

squared plus a x plus 1 squared

00:36:57

someone can tell where dc has for

00:36:59

some reason I’m not writing back why don’t you

00:37:01

write 3x square plus 2x + 1 is greater than or

00:37:04

equal to 0 so why should I write this if 3x

00:37:07

is a square

00:37:08

and forgot if 3x square plus 2x + 1

00:37:12

is equal to the expression squared so it’s

00:37:15

automatic greater than or equal to 0,

00:37:16

you know, if 3x squared plus 2x + 1

00:37:19

is written equal to the expression squared, then

00:37:21

it is greater than 0 automatically, so there are

00:37:24

only 2 conditions left, we have a system

00:37:27

of inequalities and equations, let's solve

00:37:30

the equation somehow, let's try to simplify it, I

00:37:32

remind you if this equation scares you

00:37:34

Well, seven, if seven, doesn’t

00:37:38

that also scare you? What’s scary here? Open the

00:37:43

square on the right, then

00:37:46

bring similar ones, and then look even

00:37:48

further. Here’s a way to

00:37:51

substitute a seven so as not to be scared. So

00:37:54

we have an inequality, we have

00:37:56

an equation, let’s start with the equation 3x square

00:38:00

plus 2x + 1 equals well, how to open the

00:38:06

square of the sum of three terms, you can include two

00:38:11

of the three terms

00:38:13

into one term,

00:38:16

you understand, and so we get the square

00:38:18

of the sum

00:38:19

and the square of the first expression to 4 plus

00:38:22

double the product of the first by the second

00:38:24

and plus the square of the second expression,

00:38:33

so now let’s open the brackets 3x square

00:38:37

plus the

00:38:38

routine is already starting, that is, there are only

00:38:41

one key actions here so far that has happened

00:38:43

before during all this time, this is to move on to

00:38:46

the system for everything else, well, the

00:38:47

elementary routine 2a

00:38:50

x cube plus 2x square plus

00:38:55

and square x square plus 2x + 1

00:39:01

units are mutually destroyed 2x

00:39:08

let's do it too now

00:39:10

move the 3x square to the right and

00:39:12

change the sides, that is, x 4 is further 3x

00:39:29

square and 2 x square can be

00:39:31

given like this 2 x square minus 3 x square

00:39:33

will be minus x square

00:39:36

further + 2 a x cube + a square x

00:39:42

square is equal to zero

00:39:46

4 terms here Well, grouping just suggests itself,

00:39:50

again if you, if you have a

00:39:52

gap, grouping is in what grade in

00:39:55

seventh grade does grouping probably begin,

00:39:57

in my opinion, after the abbreviated

00:39:58

multiplication formulas, if I remember correctly from the

00:40:00

main textbooks, that is, and again,

00:40:03

if you have a gap in seventh grade, then you do

00:40:05

n’t remember that there is a grouping here, you don’t

00:40:06

understand that we take out the top grouping x

00:40:08

square, it turns out x square

00:40:11

minus 1, here we enter that

00:40:15

we take out the maximum 2 x

00:40:25

2x stop, this won’t give me anything 2x yes,

00:40:30

then let’s not group it, but take out

00:40:33

x square in the whole 4 all

00:40:35

terms

00:40:36

or not you can do this grouping does

00:40:38

n’t give anything showed that it will give something no won’t

00:40:40

give x square minus 1 plus 2x +

00:40:46

and the square is equal to zero so now in

00:40:51

brackets don’t we notice anything

00:40:53

this is a niche doesn’t remind you of anything that is it

00:40:56

turns out x square multiplied by the

00:40:58

square of the sum x plus a squared

00:41:03

minus 1 well, in general this is the squared

00:41:08

difference of squares,

00:41:10

that is, it turns out x squared multiplied

00:41:13

by x plus and minus 1 multiplied by x + a +

00:41:19

1 equals zero well,

00:41:22

how to solve the equation

00:41:26

when how much do I need? multipliers are equal to zero

00:41:28

when the product is equal to zero when at

00:41:31

least one of the factors is equal, that is, we

00:41:33

say that x square is a zero or

00:41:36

is a bracket zero

00:41:44

then x is equal to minus x plus 1 and here x is

00:41:49

equal to zero or the last bracket is a zero,

00:41:55

that is, x + a + 10 iq x is equal to minus and

00:42:00

minus 1, so we got 3 x x x

00:42:06

1 x 2 x thirds, we need them all to

00:42:11

be different and so that each uri

00:42:14

opens to this, that is, to find and at

00:42:17

which excite will satisfy the

00:42:20

first inequality of the system but we have

00:42:25

there must be three different roots

00:42:27

how to require the difference look

00:42:32

at x 2 x 3

00:42:34

must not be zeros you agree

00:42:37

because if x 2 turns out to be a zero but

00:42:39

if a is equal to 1 for example then we

00:42:41

will have two roots zero and there x thirds and

00:42:44

we need three different roots that is, x 2

00:42:46

must not be a zero x 3 must not be a zero that

00:42:49

is, what we get let’s write that there

00:42:52

must be three different roots

00:43:07

and therefore castoro and she are not a zero x the

00:43:15

third must not be a zero and bloodhound 2

00:43:18

must not be x 3 x 2 not ix35

00:43:24

we will solve three conditions, that is, x 2 is not a zero and there

00:43:28

must also be x 3, also not a zero, and x 2

00:43:32

x thirds must also not be equal,

00:43:34

but because if they are equal, we

00:43:38

will have a zero and some identical

00:43:41

number there, it’s like a third-party zero it turns out

00:43:43

minus a + 1 is not a zero, that is, they are

00:43:47

one minus and minus 1 is not a zero, that

00:43:52

is, they are minus one and minus plus

00:43:57

one is not minus, but minus 1 turns out 0 and is

00:44:02

not equal to -2,

00:44:04

but any 0 multiplied by is not equal to -2 at

00:44:08

album that is, now we received an

00:44:13

additional answer that they are one

00:44:15

not minus one, this is already part of the

00:44:18

answer, we clearly got it so now

00:44:21

we need to check that each of the three

00:44:24

roots satisfies the first inequality

00:44:27

in the system,

00:44:28

let's write each of the three roots

00:44:40

they are already different and this one, as it were, was made so

00:44:45

that they were different, they are different

00:44:47

if they are 1 and not -1, if they are -1 and 1,

00:44:50

they are exactly different, all three are a fact, so

00:44:53

each of the three roots must

00:44:57

satisfy the first inequality; the

00:44:59

system satisfies and the ones, well,

00:45:07

we substitute everything, which means zero minus x plus 1

00:45:10

under x and minus a minus 1 and look at

00:45:13

what a they will actually netware

00:45:15

satisfy this inequality so

00:45:18

the first

00:45:20

with x first equal zeros second

00:45:29

with x second equal descendant old was

00:45:34

so 2 was minus x plus 1 and 3 with x 3

00:45:43

minus a minus 1 and so under

00:45:46

the inequality we substitute he she at the top x

00:45:48

square plus a x + 1 greater than or equal to 0

00:45:51

it turns out 0 squared plus a multiplied

00:45:59

by 0 plus 1 greater than or equal to 0 3 1 and I

00:46:03

’ll rewrite mine so that you don’t look there every time

00:46:05

x squared plus x plus 1 greater

00:46:10

equals 0 and substitute a zero and these

00:46:12

two are also the values of 0 squared plus and

00:46:15

multiplied by 0 plus 1 large equals 0

00:46:18

let’s see at what conditions

00:46:23

the inequality will be satisfied so it turns out 0

00:46:26

and large equals minus 10 multiplied by

00:46:32

any number greater than and equal to -1 that

00:46:34

is, or maybe any

00:46:39

here we substitute instead of exe for minus a

00:46:42

minus 1 it turns out minus a minus one

00:46:44

moment a + 1 more precisely squared plus a

00:46:49

multiplied by minus x plus 1 plus 1 greater than

00:46:55

or equal to 0 so we open the square of

00:46:58

the difference, that is, one minus 2 plus 1

00:47:01

minus x square plus a square more precisely on

00:47:06

so minus square + a + 1 is greater than or

00:47:11

equal to 0 square and leave it turns out a is

00:47:16

less than or equal to topic 2

00:47:20

so that is, when a is less than or equal to 2 x 2

00:47:24

satisfies the inequality in the system and x

00:47:27

3 we substitute that is x square plus bx

00:47:31

plus one again here we substitute

00:47:32

all this was the first inequality of the system

00:47:36

at the beginning it turns out minus a minus 1

00:47:39

squared plus a multiplied by minus a

00:47:42

minus 1 plus 1 greater than or equal to 0

00:47:49

since it’s a square, but let’s put the minus out of the bracket and the

00:47:51

square will kill this minus, it turns out

00:47:53

square plus 2 plus 1 minus x square

00:47:56

minus x plus 1 is greater than or equal to 0

00:48:00

square and they leave a greater than equal to -2,

00:48:04

that is, when a is greater than or equal to -2 the

00:48:08

third root satisfies the inequality

00:48:11

at the beginning,

00:48:14

that is, for any and the first root

00:48:18

satisfies the inequality began directly

00:48:20

less than 2 2 root 2 charge

00:48:23

inequality began and for a greater faith of

00:48:24

the night minus 2 3 potassium google opens

00:48:27

the inequality at the beginning while they are 1 and not

00:48:30

-1 that is, what are the restrictions for we

00:48:32

get we get that a must be

00:48:40

one and must not be minus

00:48:43

one it is necessary that the roots be

00:48:46

different

00:48:47

and must be less than or equal to 2 it is

00:48:49

necessary that the second root fits the

00:48:52

inequality in system 1 and a is greater

00:48:56

than -2 this is necessary so that the third root

00:48:58

fits the inequality first in the system

00:49:01

we find the intersection of these restrictions for

00:49:04

and this will be our answer you can on the number

00:49:08

line you can in principle find

00:49:10

the intersection in a month give on the number line that

00:49:12

is we put -2 shaded

00:49:15

-1 empty one empty

00:49:18

barely they are painted one not -1 you need them

00:49:23

you carried calls them and less is equal to 2

00:49:26

and more is equal to -2 where 2 shadings are there is

00:49:33

the answer so the answer is in the

00:49:35

VKontakte group lead no thanks for watching

00:49:40

bye

Description:

Что такое параметр (разберём теорию). Привет, меня зовут Евгений Пифагор, и я готовлю к ЕГЭ по математике более 10 лет. Решим задачи с параметром и разберём методы их решения. Видео будет полезно тем, кто изучает параметры с нуля. Разберём графический метод (для параметра с модулем) и аналитический метод решения (для параметра с корнем) на примерах задач уровня ЕГЭ. Разберём виды условий задач с параметром: - найти все значения параметра а при которых ... - при всех а ... 👍 ССЫЛКИ: Подпишитесь: https://www.youtube.com/c/pifagor1?sub_confirmation=1 VK группа: https://vk.com/shkolapifagora Видеокурсы: https://vk.com/market-40691695 Insta: https://www.facebook.com/unsupportedbrowser 🔥 ТАЙМКОДЫ: 00:00 – определение параметра 00:15 – пример на понимание что такое параметр 03:35 – ДЗ#1 03:55 – знакомимся с десмосом 09:20 – ДЗ#2 09:37 – задача уровня ЕГЭ, решённая с помощью десмоса 19:35 – ДЗ#3 19:43 – что значит решить уравнение (неравенство) при всех а 24:07 – ДЗ#4 24:20 – виды условий параметров 29:02 – сколько баллов на ЕГЭ может принести параметр 31:35 – изучаются ли параметры в школе? 33:07 – способ понять условие задачи с параметром (способ с семёркой) 33:24 – задача уровня ЕГЭ, решённая с помощью способа с семёркой 50:45 – ДЗ

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