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|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
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Ta thấy:
\(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x-3\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x+1\right|-\left|x-3\right|\ge0\)
Mà \(\left|x+1\right|-\left|x-3\right|\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+1\right|=0\\\left|x-3\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
Không chắc :))
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