The absolute value of 2x 1 plus the absolute value of x 2 is simplified?

This question is going to be discussed because we don't get to the absolute value sign that the number is exactly negative. The absolute value of a positive number is herself, and the absolute value of a negative number is her opposite.

This question is divided by 1, 2, and 2.

Specific process: (the explanation in parentheses is not written when solving the problem).

When x 2 (both absolute values are positive) 2x-1 + x-2 = 3x-3 when 1 2< x<2 (the former is positive, the latter is negative), 2x-1-(x-2)=x+1, when x<1 2 (both are negative, so the absolute value sign is removed, both equations become the original opposite) -(2x-1)-(x-2)=3-3x< p>

Dig your question is too difficult to do, there should be several answers: 1: When 2x-1>0 x-2>0 is 2x-1+x-2=3x-3 or =3(x-1) 2: When 2x-1<0 x-2<0 is 1-2x+2-x=3-3x or 3(1-x) 3:

When 2x-1>0 x-2<0. Let's count it myself, too much, a total of 6 cases, hehe, there are 2 very important situations, 1 is when 2x-1 and so on, 0 2 is x-2=0, hehe, there are a total of 6 situations, I think so.

1.When x>=2, the original formula is 3x-3

2.When -1 2<=x<2, the original formula is 1+x

3.When x<1 2, the original 3-3x

This question should be discussed, where x-2 = 0 and 2x-1 = 0, i.e. x=2 or x=1 21When x>=2, the original formula is 3x-3

2.When -1 2<=x<2, the original formula is 1+x

3.When x<1 2, the original 3-3x

When x 2, reduce to 3x-3, when x 1 2, simplify to 3-3x, and when 1 2 x 2, reduce to x+1.

|x-2|-x+1|=?

1).When x-2>=0, that is, |x-2|=x-2, so the original =|x-2-x+1|=|1|=1 file to do [x>=2].

2).When x-2>=0, that is, |x-2|=-x-2)=-x+2, so the original =|-x+2-x+1|=|2x+3|

At this time, it is also necessary to divide, if -2x+3>=0, the formula = -2x+3 [x<=if -2x+3<0, the empty stupid sedan formula =-(2x+3)=2x-3 [

x-2=0，x=2；x+1=0，x=-1.

x=-1 and x=2 divide the number axis into three segments: (-1], (1,2], (2, difference and +) virtual marking.

When x (-1], |x-2|-|x+1|=-x-2）-（x+1））=x+2+x+1=3,；

When x (-1,2], Kuanxiang |x-2|-|x+1|=-x-2）-（x+1）=-x+2-x-1=-2x+1；

When x (2,+, |x-2|-|x+1|=x-2-（x+1）=x-2-x-1=-3。

x-2|+|2-x|

2|x-2|

When x 2, the original formula = 2 (2-x) = 4-2x, when x > the other wheel pose of the tung eggplant, the original vertical formula = 2 (x-2) = 2x-4

x+.2 of absolute. How about the value. Simplify.

x+2| =x| +2

Here's the answer: the absolute value of +2 is simplified like this, first with one plus one and then with one plus two.

The original base Li Sen question is to fight the mu|x+2|-|x+1|Is it?

Yes words. Will|x+2|Rewrite the disturbance of the forest into |x+1+1|

So |x+1+1|-|x+1|=|x+1|+1-|x+1|=1

When x > 2, 2 is an absolute value of x.

x-2When x=2, the absolute value of 2-x = 0

When < 2 is the filial brother, the value of 2 x is 2 x = 2 x .

When x>2, the absolute value of 2- = -2

When x = 2, the absolute value of 2-x = 0

When <2, the pair value of the Absolute Void Stove of the Nest 2-x is infiltrated = 2-x.

2 a x when 2 a x hungry bridge 0, that is, x 2, 丨 years old 2 a x丨 2 a x, when x 2, 丨 2 a x丨 0, when the rotten x 2, 丨 2 x x a 2

When 2-x=0, 丨2- |2- when <2|2- Chaichai丨=2-x, when the filial piety >2, read and judge, 丨2- |2

2-x When x Tong Duan Xun 2 burns friends, 2-x round this = 2-x.

When x 2, 2-x = x-2

Let x+1=0, then.

If x=-1 makes x+2=0, then x=-2

When the source x<-2: baix+1<0, x+2<0 then the original =-(x+1)+[x+2)]=-x-1-x-2=-2x-3 When -2 x -1: x+1 0, x+2 0 then the original =-(x+1)+(x+2)=-x-1+x+2=1 When dux>-1:

x+1>0, x+2>0 then the original = x+1+x+2=2x+3

When x 0, the absolute value of x+1 plus the absolute value of x+2 is reduced to 3;

When x<0, the absolute value of x+1 plus the absolute value of x+2 is reduced to 3-2x;

When x>0, the absolute value of x+1 plus the absolute value of x+2 is reduced to 2x+3;

If x+1=0, then x=-1

Let x+2=0, then x=-2

When baix<-2: x+1<0, dux+2<0, then the original zhi formula =-(x+1)+[x+2)]=-x-1-x-2=-2x-3

When -2 daox -1: x+1 0, x+2 0

Then the original formula =-(x+1)+(x+2)=-x-1+x+2=1 When x>-1: x+1>0, x+2>0

then the original formula = x+1+x+2=2x+3

|y=|zhix+2|+|x-2|+|x-1|, daox 2, y=x+2+x-2+x-1=3x-1 5, 1 x 2, y=x+2-x+2+x-1=x+3,4 y 5,-2 x 1,y=x+2-x+2-x+1=-x+5,47, in summary, x=1, there is a minimum value y=4

Discuss on a case-by-case basis. , the value can be x<=-1 and -11, respectively. Analyse.

Take the first one as an example, when x<=-1, |x+1|=-x-1,|x-2|=2-x,|2x-1|=1-2x, therefore, the original formula =-x-1+2-x-(1-2x)=2.

In other cases, if the number in the absolute value is calculated to be a negative number, the simplification process is to multiply the number in the absolute value by a negative sign, and then perform other operations OK, if the number in the absolute value is a positive number, then directly remove the absolute value sign is OK.

Friends, it's important to master the method!

Sorry to write out a minus sign.

The answer is yes hahaha.

It's too hard to dig up your problems.

There must be several answers.

1: When 2x-1>0

x-2>0.

That is, 2x-1+x-2=3x-3 or =3(x-1)2: when 2x-1<0

x-2<0

That is, 1-2x+2-x=3-3x or 3(1-x)3: when 2x-1>0

x-2>0. The following self-calculation is too much, a total of 6 situations.

Hehe, there are 2 very important situations.

1 is when 2x-1 and so on 0

2 is x-2=0

Oh, there are a total of 6 situations.

I think so.

Suppose x is greater than 2, then.

2x-1 is less than 2-x

3x less than 3x less than 1

Non-conformance 2x-1 is less than x-2