There are 2 inequality questions in the first year of junior high school, and one inequality questio

In the first problem, move m to the right is 2x>m-3, and the solution set obtained is x>-2, that is, 2x>-4, compare the two formulas, because the solution x is the same, so the value on the right side of the two equations is the same, m-3=-4, m=-1.

The second question, diagonal multiplication: 3 (a+x)>=2(2x-1), remove parentheses, 3a+3x>=4x-2, move left and right, 3a+2>=4x-3x, that is, 3a+2>=x, that is, x<=3a+2, compared with x<=8, know that 3a+2=8, so a=2, the input method switch is really tired.

2x>m-3

x>(m-3)/2

m-3)/2=-2m=-1

3a+3x>=4x-2

3a+2>=x

The answer is written as you add ...

3a+2=8a=2

1,x>(m-3) 2=-2 so m=-1

2, x is less than or equal to 3a+2=8 so a=2

2x>m-3

x>(m-3)/2

x>-2 again

So (m-3) 2=-2

m=-1a+x) 2 >= (2x-1) 33a+3x>=4x-2

x<=3a+2

Because x<=8

So 3a+2=8a=2

Solution: Combination of numbers and shapes.

Specific steps: Solution: Because of 3x-a 0

So to solve the inequality x a 3, because it has only 3 positive integer solutions, the three integer solutions must be 1, 2, 3 to draw the number axis, as follows:

To meet the conditions, 3 A 3 4 is required, and the solution is: 9 A 12, that is, the value range of a is 9 A 12

｛x-5}-＜1

Note: {} means absolute).

Answer as follows: x<-2 3, 5-x+2x+3 1

Solution x<-7

At x<5, 5-x-2x-3 1

The solution is x> wide and smooth 1 3

X-5-2x-3 at x>5

Solution x>-9

To sum up: x<-7

or 1 35 positive rational number a1 is an approximation of the root number 3, let a2=2 (a1+1), and verify that the root number 3 is between a1 and a2.

a2=2 (a1+1)>0,a1>0, then a1 a2=a1 [2 (a1+1)]=a1(a1+1)] 2>=[root number 3(root number 3+1)] 2>1

So there are a1>a2, and from the positive rational number a1 is an approximation of the root number 3, 0 root number 3 (proved by the right endpoint).

Inequality +100 has (

Group integer solutions. It can be approximated that the inequality expressed by Shen Hela refers to the calculation of the number of all points in a rectangular area surrounded by four straight lines with intercepts of 100 in the plane Cartesian coordinate system (integer unbeat), and the value of x can vary from 0 to 99, and each x value will correspond to two.

y, so there is 99*2 198

By the way, there will be a reward.

Set: x children, y apples.

then y-3x=8

0<=y-5(x-1)<3

Solution: x=6

y=26 should be x, y must be an integer.

There are x children with y apples.

3x+8=y

y-(x-1)*5<3

Or: 3x+8-(x-1)*5<3

Set: x children, y apples.

3x+8=y

y-5（x-1）<3

x>5y>23

Solution: If there are x children, then there are (3x+8) apples from the meaning of the question.

0＜（3x+8）-5（x-1）

3x+8）-5（x-1）＜3

Solution 5 x

x is a positive integer x=6

So there are 6 children with 3 * 6 + 8 = 26 apples.

1、|x-2|+|x-3|≥m

Solution: i.e., the distance from x to 2 + the distance from x to 3 is m

When m<=1, the solution set of x is the whole real number.

When m>1.

x 3, 2x-5 m x (5+m) 2x<=2, 5-2x m x<=(5-m) 2, i.e. x (5+m) 2 or x<=(5-m) 22, |x+3|-|x-3||3 Square on both sides.

2x^2+9-2|x^2-9|>0

x 2-9>=0, 27>0 x>=3 or bright-burning x<=-3x 2-9<0, x 2>9 4 3 21, x>m-2m=1, x does not exist

When m<1, x1, x>m-2 m-2>=3 2m>1, m>=7 2

m>=7/2

m<1, x=7 2 or m<=1 2

(1)3x-y=k+4 (2)

14x=14

x = 1 substituted (2).

3-y=k+4

y=-k-1

x-y=1+k+1=k+2

k|<3

318 7a>28 a>4

The value range of 4a is (4,6).