SOLUTION INEQUALITY X 2 2X 1 1

Solving inequalities|x+2|-|2x-1|≥1

Solution: x+2 -2 x-1 2 1

When x -2, there is -(x+2)+2(x-1 2)=x-3 1, so x4 (>2, rounded, this paragraph has no solution).

When -2 x 1 2 there is x+2+2(x-1 2)=3x+1 1, so we get x 0, that is, 0 x 1 2....1) The solution to this paragraph.

When x 1 2 there is x+2-2 (x-1 2) = -x+3 1, so x 2 is obtained, i.e. 1 2 x 2....2) The solution to this paragraph.

1) (2) = is the set of solutions to the original inequality.

..When x -2, the inequality is.

x+2|-|2x-1|≥1

x-2-（－2x+1）≥1

x 4 x -2 is not true when x -2 is at this time.

When -2 x 1 2, the inequality is.

x+2-（－2x+1）≥1

x≥0∵-2≦x＜1/2

0≦x＜1/2

When x 1 2, the inequality is.

x+2-（2x-1）≥1

x≦3∵x≧1/2

1/2≦x≦3

Combined 0 x 3

It would be nice to have a compartmental discussion.

When x is greater than -2, remove the unequal sign then -x+1 1 solves x less than or equal to 0, so x takes the value from -2 to 0

When x is greater than you discuss it.

When x is -2 to the point where it is, you go to the absolute value and discuss it, just notice the existence of emptiness.

Categorical discussion, divided into (- 2) [2, 1 2], 1 2, + three cases, after removing the absolute value.

The original inequality is equivalent to:

2x-1) (x+2) 1 Qi 0

2x-1) (x+2) (x+2) (x+2) 0 x-3) (x+2) 0

x-3)(x+2) 0 and x≠-2

2＜x≤3

The solution set of the equation of the original tomb and the tomb is.

This is an absolute inequality, so consider removing the absolute value symbol. It is known by the definition of absolute value: when x

Analysis: (1) x Zhengfan 2, then 2 x +1+ x -2 4 x Xiao Meng 22)- x 2, then 2 x +1+2- x 4 1 x 2

3) x-then-2 x -1- x +2 4 x -1To sum up, the solution is as follows. .

Method 1: (1) When x 1, the inequality can be reduced to 1-x+2-x 2 to obtain x

2) When pretending to be 1 x 2, the inequality is x-1+2-x 2 and there is no solution.

3) When x 2, the inequality is x-1+x-2 2, and the solution of x is combined to obtain x or x

Method 2: |x-1|+|2-x|Represents the sum of distances between (x,0) and (1,0) and (2,0).

For the sum of distances to be greater than 2, x or x must be satisfied.

When x<1, |x-1|+|x-2|=1-x+2-x=3-2x gives 3-2x<2, and the solution gives x>1 2

i.e., when 1 2=2, |x-1|+|x-2|=x-1+x-2=2x-3 gives 2x-3<2, and the solution gives x<5 2

That is: 2<=x<5 2

In summary, the value of x can range from 1 to 2

The solution to this inequality:

1.Divided into three cases< x>1, x2, 1 x 2, it becomes a group of three inequalities 2Make use of geometric meaning:

It is not difficult to find the sum of x and |, which represents a point on the number line where the sum of distances to 1 and 2 is less than 2x-1|+|x-2|2 Therefore, the set of solutions of the original inequality is: (,

1. x is less than or equal to 1

1-x+2-x<2

3-2x<2

x is greater than 1 2

So x is greater than 1 2 less than or equal to 1

2. x is greater than or equal to 1 and less than or equal to 2

1-x+x-2<2

Constant so x is greater than or equal to 1 and less than or equal to 2

3. x is greater than or equal to 2

x-1+x-2<2

2x-3<2

x<5/2

So x is greater than or equal to 2 and less than 5 2

Composite x is greater than 1 2 less than

When x is greater than 1 and less than 2, remove the absolute value before the unchanged, and add a negative sign after it to solve x as an empty set. When x is less than 1, the absolute value of both sides is removed and the negative sign is added at the same time to solve the solution of x is greater than half of the negative. When x is greater than 2, the absolute value remains unchanged, and the solution is less than five-twoths.

So x is greater than one-half and less than five-two.

Solution: |2x-1|+|x-2|4 rules.

x<1/2;-(2x-1)-(x-2)>4 or {1 2 x 2; +(2x-1)-(x-2)>4 or {x>2; +(2x-1)+(x-2)>4③

By x<-1 3

By De X

By x>7 3

From or or get x<-1 3 or x>7 3.

1+2x=0, x=-1/2 2+x=0, x=-2

When x>-1 2, the original formula = 1+2x-2-x-2<0 , x<3 is -1 2-1 , and x<-2, so there is no solution.

In summary, -1 2

:|1+2x|-|2+x|<2

1+2x|<|2+x|+2

Take the square to get.

4x²+4x+1<4+x²+4x+4+4|2+x|Move 3x <7+4|2+x|

3x²-7<4|2+x|

Obviously, if 3x -7<0, the inequality must hold.

At this time, x <7 3 is solved to square again when the root number 21 3=0.

9x^4-24x²+49<16（x²+4x+4）9x^4-40x^2-64x-15<0