X X 1 X 2 X 3 8. Please help, thank you!

x(x+1)(x+2)(x+3) 8, find the range of x values.

Solution: [x(x+3)][x+1)(x+2)]-8<0x +3x)(x +3x+2)-8=(x +3x) +2(x +3x)-8=(x +3x+4)(x +3x-2)<0

Since the discriminant formula of x +3x+4 δ=9-16=-7<0, for any x constant x +3x+4>0, the original inequality is the same solution as x +3x-2<0, and the solution of this inequality is (-3- 17) 2

Let k=x+3 2

x(x+1)(x+2)(x+3)=(k^2-1/4)(k^2-9/4)

Let k 2-5 4=m again

m^2-9<0

Solve m, and then push it backwards.

The answer is too lazy to count, troublesome.

Since x>3, x-3>0(x 2-3x+1) (x-3)=[x(x-3)+1] (x-3)=x+1 (x-3)=x-3+1 (x-3)+3> fast clearing number = 2 [(x-3)*1 (x-3)]+3=2+3=5 if and only if x-3=1 (x-3), from the mu to x-3>0, so x-3=1, x=4 so when x>3, (x 2-3x+1) (x-3) The range of positive groups is [5,+

|x+1|=x+1, the number in the absolute value is equal to itself, so this number is greater than or equal to 0, that is.

x+1≥0,x≥-1

Again|3x+2|=-3x-2=-(3x+2), the number in the absolute value is finally equal to its opposite number, indicating that the number of the sedan car is less than or equal to 0 in the Wang Fan calendar

3x+2≤0,x≤-2/3

In summary. 1≤x≤-2/3

The key idea of this problem is to find a way to remove the absolute value symbol.

There are three cases: 1) When x is in [1, directly remove the absolute value sign x-1+2+x=3,x=1 is true.

2) When x is at [-2,1), we get x+2+1-x=3, and we get x belongs to [-2,1).

3) When x belongs to (- 2), 1-x-2-x=3 is solved to obtain x=2 (rounded). The value of x can be -2,1

Greater than or equal to 0 and less than or equal to 3

This kind of topic should be discussed on a case-by-case basis, first find out the critical value, that is, x-1=0, 2+x=0 x are 1, -2 respectively. Then we will discuss it in three situations.

According to the title, (x-2) and (x-3) have the same sign. That is, (x-2) multiplied by (x-3) is greater than or equal to zero. Push x is not less than 3, or x is not greater than 2.

(x-2)+(x-3)|=x-2|+|x-3|Because the equation is greater than 0 on both sides and left

So the two sides are squared.

x-2)²+2（x-2)(x-3)+(x-3)²=2|x-2||x-3|+(x-3)²+x-2)²

Simplify(x-2)(x-3)=|x-2||x-3|So x<2 and x<3 or x>2 and x>3

Get x<2 or x>3

x+1) -1 3<(3-2x) -1 31, when Tanwu loses x+1>0 and 3-2x> let 0 i.e., -13-2x, the solution is x>2 3 i.e., 2 30 and 3-2x<0, i.e., x>3 2, there is (x+1) 1 3 (3-2x) 1 3 The positive number is less than the negative number, and there is no solution.

3. When x+1<0 and 3-2x<0, x<-1 and x> orange Wu 3 2 have no solution.

4. When (x+1) < 0 and (3-2x) >0, x<-1 is combined with x<-1 or 2 3

The original formula is equivalent to 1 (x+1)<1 (3-2x)1, when (x+1)>0 and (3-2x)>0 i.e., -13-2x, the solution gives x>2 3 i.e., defeat 2 30 and (3-2x)<0, i.e., x> field 3 2, and x, which satisfies the condition, does not exist.

3. When (x+1)<0 and (3-2x)<0 and (x+1)<0 and (3-2x)>0, there is no x that satisfies the conditions

To sum up, the value range of x is 2 3

(x+1) -1 3 "Early Age Hood (3-2x) -1 3(x+1)> (3-2x)

Lu muffled his eyes 3x>2

Get x>2 3

To discuss 1, when x 0, the original equation is: -x+3=3-x So 3=3 is the identity. So the conditions are met.

2. When 0 x 3, the original equation is: x+3=3-x The solution of x=0 does not meet the condition, and it is rounded.

3. When x 3, the original equation is early, x+3=x-3, and the equation is 3=-3 is not true

I hope I can help you and have a great day.