Ta có\(\left|x+1\right|-\left|x-3\right|=0\)
   \(\Rightarrow\orbr{\begin{cases}\left(x+1\right)-\left(x-3\right)=0\\\left(x+1\right)+\left(x-3\right)=0\end{cases}}\)

Nếu \(\left(x+1\right)-\left(x-3\right)=0\)
   \(\Rightarrow x+1=x-3\)\(\Rightarrow x=x-4\)
   \(\Rightarrow x\)không tồn tại.

Nếu \(\left(x+1\right)+\left(x-3\right)=0\)
   \(\Rightarrow x+1=-\left(x-3\right)=3-x\)
   \(\Rightarrow2\left(x+1\right)=\left(x+1\right)+\left(x+1\right)=\left(x+1\right)+\left(3-x\right)=x+1+3-x=4\)
   \(\Rightarrow x+1=\frac{4}{2}=2\)
   \(\Rightarrow x=2-1=1\)

Vậy \(x=2\)

lx+1l - lx-3l=0

 x=2;x=6

Thay số:l2+1l - l6 - 3l=0

  Vậy x=2;x=3

í lộn x=6 chứ

|x+1|-|x-8|=0 (1)

Xét x<-1 => |x+1|=-x-1, |x-8|=8-x

(1) <=> (-x-1)-(8-x)=0 

<=> -x-1-8+x=0 

<=> -7=0 (vô lí)

Xét -1=<x<8 => |x+1|=x+1, |x-8|=8-x 

(1) <=> (x+1)-(8-x)=0 

<=> x+1-8+x=0 

<=> 2x-7=0 

<=> x=7/2 (thỏa mãn) 

Xét x>=8 => |x+1|=x+1, |x-8|=x-8 

(1) <=> (x+1)-(x-8)=0 

<=> x+1-x+8=0 

<=> 9=0 vô lí

Vậy x=7/2

=>x-2=0; x+y=0; 2x-2z=0

=>x=0; y=0; z=0