Solve the inequality 3X 2 2X 3

x<-3 2, -3x-2>-2x-3, get x<1, and hold at the same time, so x<-3 2.

3 2<=x<=-2 3, -3x-2>2x+3, we get x<-1, so, -3 2<=x<-1.

When x>=-2 3, 3x+2>2x+3, we get x>1, so x>1.

Three-stage synthesis yields x<-1 or x>1.

Assuming that x is 0 and 2 is greater than 3 is not true.

Assuming x is positive, then 3x+2 is greater than 2x+3 x is greater than 1

Suppose x is negative, then 2-3x is greater than 3-2x, -5x is greater than 1, -x is greater than 1, x is less than -1, so x is greater than 1 and x is less than -1

Both sides are squared at the same time, resulting in 9x 2 12x 4 4x 2+12x+9, i.e. x 2>1

Solve x>1 or x<-1

Write the standard point yourself)

The squares of both sides give (3x+2) 2>(2x+3) 2 shifts and subtract them to get (x-1)(x+1)>0

There are two scenarios.

1. x-1> and x+1>0 get x>1

2. x-1<0 and x+1<0 get x<-1

Summary: The solution of inequality is.

Both sides are squared, got.

9x 2+12x+4>4x 2+12x+9 i.e.

5x^2>5

x^2>1

x>1 or x<-1

It's done, and those who have money have a money field.

2x 2+x+3<0 [Unsolved].

When 2x 2+x+3=0

b^2-4ac=1-24=-23<0

2x^2+x+3>0

When 2x 2+x+3=0

b 2-4ac=1-24=-23 "Slow comma key 0 scramble finger pants x is an arbitrary real number].

Because the image of the function f(x)=2x +x+3 is all above the x-axis, it is absolutely rotten.

The solution set of 2x +x+3>0 is r, and the two omissions of the first cavity draft are empty sets.

To facilitate your understanding.

So the inequality is decomposed into two inequalities, which can be decomposed by the definition of absolute values.

12.-(3-2x)>1

Solving for the first inequality yields: x<1

Solving for the second inequality yields: x>2

Since the two solutions do not intersect.

So the answer is "or".

That is: x<1 or x>2

Using the piecewise discussion method, let the equation in the two absolute values be equal to 0 respectively, and find the two values of x as the segmentation points, in this problem, x=2 and x=-5 are the segmentation points. Then it is divided into three segments according to the segmentation point, which is: 1)x<-5;2）-52；

1) x<-5, the original inequality becomes 2-x-x-5>3, that is, x<-3, so the intersection is x<-5;

2)-53, the constant holds, the intersection gets -52, the original inequality becomes x-2+x+5>3, that is, x>0, and the intersection gets x>2;

In the above three cases, x is not equal to -5 or x is not equal to 2. In addition, if there are three absolute values, the same method can be used. Thank you!

Number line,|x-2|+|x+5|>3，，|x-2|denotes the distance from the point to the point 2, |x+5|denotes the distance from point to point -5, inequality |x-2|+|x+5|The solution of >3 is the set of all points represented in the number axis if the sum of the distance from a point to 2 and the distance from a point to -5 is greater than 3.

Solution Inequality Group:

3x-1>2x+1, then x>2

2x>8 then x>4

So: the set of solutions for the group of solution inequalities is:

3x-1>2x+1

3x-2x>1+1

x>22x>8

x>4 can ask if you don't understand.

Satisfied, thank you.