Remove the absolute value of the problem, simplify 丨丨x 1丨 3丨 丨3x 1丨

When x>=1, 丨x-1丨=x-1, 丨丨x-1丨-3丨+丨3x-1丨=丨x-4丨+丨3x-1丨.

If x>=4, the original formula = x-4 + 3x-1 = 4x-5 if 1 = if 1 3=

Here's how! x-1 0, get x=1

3x-1 0, get x=1 3

丨x-1丨-3=0. x = 4 or -2

Segmentation to remove absolute values and simplify.

x<-2 -x+1-3-3x+1=-4x-1x>=-2 x<1 3 x-1+3-3x+1=-2x+3x>=1 3 x<1 x-1+3+3x-1=4x+1x>=1 x<4 3-x+1+3x-1=2x+3x>=4 x-1-3+3x-1=4x-5 is not difficult, but it is more troublesome, and every paragraph must be discussed!

When x 4.

x-4+3x-1=4x-5

When 4 x 1.

丨x-1-3丨+3x-1=4-x+3x-1=2x+3 when -2 x 1.

丨1-x-3丨+1-3x=丨-x-2丨+1-3x=2+x+1-3x=3-2x

When x -2.

丨1-x-3丨+1-3x=-2-x+1-3x=-1-4x

1 x 3, so x-3 0, |x-3|=-x-3）=3-x；Base type x-1 0,|x-1|=x-1

Therefore, the original guess is rolled = x-1 + 3-x = 2

3x+1|+|2x-1|=

To simplify this kind of problem is to remove the absolute or dust deficit symbol, and the method is generally to use the discussion method. You can draw a number axis, take 3x+1=0, 2x-1=0, that is, x=-1 brother 3 and x=1 2, they divide the number axis into three parts, and discuss x. in these three parts

1, when x0, the original = |3x+1|+|2x-1|=3x+1+2x-1=5x

If Knott's 1 x 3, the absolute value of x 3 on the Reduced x 1 pair is (2).

When 1 x 3 is not taken, x-1>0, |x-1|=x-1x-3

x -1, :|x|+|x+1|-|x-1|=-x-x-1+x-1=-x-2；

When 1 x 0, the original formula =-x+x+1+x-1=x;

When 0 x 1, the original missing hand = x + x + 1 + x - 1 = 3x;

When x is disguised with 1, the original macro laugh = x+x+1-x+1=x+2