Question

\(\sqrt{-2x+6}=x-1\)

Answer 1

ĐKXĐ: \(-2x+6\ge0\Leftrightarrow-2x\ge-6\Leftrightarrow x\le3\)

Ta có: \(\sqrt{-2x+6}=x-1\)

\(\Leftrightarrow-2x+6=\left(x-1\right)^2\)

\(\Leftrightarrow-2x+6=x^2-2x+1\)

\(\Leftrightarrow-2x+6-x^2+2x-1=0\)

\(\Leftrightarrow-x^2+5=0\)

\(\Leftrightarrow x^2=5\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}\left(nhận\right)\\x=-\sqrt{5}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{\sqrt{5};-\sqrt{5}\right\}\)