`a, x - 1 = x^2 - 1`
`-> x^2 - x = 0`
`-> x(x-1) = 0`
`-> x = 0` hoặc `x = 1`
`b, 2x - 1 = x - 1`
`-> x = 2`
a, Đk: \(x^2-1\ge0\Leftrightarrow\left(x-1\right)\left(x+1\right)\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x+1\ge0\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x-1\le0\\x+1\le0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)
\(\sqrt{x^2-2x+1}=x^2-1\)
\(\Leftrightarrow\left|x-1\right|=x^2-1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=x^2-1\\x-1=1-x^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x=0\\x^2+x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x-1\right)=0\\\left(x-1\right)\left(x+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=1\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)
- Vậy \(S=\left\{1;-2\right\}\)
b, Đk: \(x-1\ge0\Leftrightarrow x\ge1\)
\(\sqrt{4x^2-4x+1}=x-1\)
\(\Leftrightarrow\left|2x-1\right|=x-1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x-1\\2x-1=1-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)
- Vậy S=∅
