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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 \(x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
|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
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,3x-31=-40\Rightarrow3x=-9\Rightarrow x=-3\)
\(b,-3x+37=\left(-4\right)^2\Rightarrow-3x=-21\Rightarrow x=7\)
\(c,\left|2x+7\right|=5\)
\(\Rightarrow\left\{{}\begin{matrix}2x+7=5\Rightarrow x=-1\\2x+7=-5\Rightarrow x=-6\end{matrix}\right.\)
\(d,-x+21=15+2x\Rightarrow3x=6\Rightarrow x=2\)
a) Ta có: 3x-31=-40
\(\Leftrightarrow3x=-9\)
hay x=-3
Vậy: x=-3
b) Ta có: \(-3x+37=\left(-4\right)^2\)
\(\Leftrightarrow-3x+37=16\)
\(\Leftrightarrow-3x=16-37=-21\)
hay x=7
Vậy: x=7
![](https://rs.olm.vn/images/avt/0.png?1311)