Question

\(x^2+x-1=\left(x+2\right)\sqrt{x^2-2x+2}\)

Answer 1

\(\Leftrightarrow x^2-2x+2-\left(x+2\right)\sqrt{x^2-2x+2}+3x-3=0\)

Đặt \(\sqrt{x^2-2x+2}=t>0\)

\(\Leftrightarrow t^2-\left(x+2\right)t+3x-3=0\)

\(\Delta=\left(x+2\right)^2-12\left(x-1\right)=x^4-8x+16=\left(x-4\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\frac{x+2-\left(x-4\right)}{2}=3\\t=\frac{x+2+x-4}{2}=x-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-2x+2}=3\\\sqrt{x^2-2x+2}=x-1\left(x\ge1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x+2=9\\x^2-2x+2=x^2-2x+1\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow x^2-2x-7=0\)