Bài 3:
a) Ta có: 2x-6=x+1
\(\Leftrightarrow2x-6-x-1=0\)
\(\Leftrightarrow x-7=0\)
hay x=7
Vậy: S={7}
b) Ta có: x-2=3x+1
\(\Leftrightarrow x-2-3x-1=0\)
\(\Leftrightarrow-2x-3=0\)
\(\Leftrightarrow-2x=3\)
hay \(x=\frac{-3}{2}\)
Vậy: \(S=\left\{\frac{-3}{2}\right\}\)
c) Ta có: |x+1|=5
\(\Leftrightarrow\left[{}\begin{matrix}x+1=5\\x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)
Vậy: S={-6;4}
d) Ta có: |x-8|=2x+5(*)
Trường hợp 1: \(x\ge8\)
(*)\(\Leftrightarrow x-8=2x+5\)
\(\Leftrightarrow x-8-2x-5=0\)
\(\Leftrightarrow-x-13=0\)
\(\Leftrightarrow-x=13\)
hay x=-13(loại)
Trường hợp 2: x<8
(*)\(\Leftrightarrow8-x=2x+5\)
\(\Leftrightarrow8-x-2x-5=0\)
\(\Leftrightarrow3-3x=0\)
\(\Leftrightarrow3x=3\)
hay x=1(nhận)
Vậy: S={1}
