Question

Tìm x, biết

\(\sqrt{x^2-2x+4}=2x-2\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\4x^2-8x+4=x^2-2x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x^2-6x=0\\x\ge1\end{matrix}\right.\Leftrightarrow x=2\)

Answer 2

\(\sqrt{x^2-2x+4}=2x-2\)             Đk: \(2x-2\ge0\Rightarrow x\ge1\)

\(\Rightarrow x^2-2x+4=\left(2x-2\right)^2\)

\(\Rightarrow x^2-2x+4=4x^2-8x+4\)

\(\Rightarrow3x^2-6x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)