Question

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

Answer 1

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

Đặt \(\sqrt{x^2-2x+2}=a>0\)

\(\Rightarrow a^2-\left(x+2\right)a+3x-3=0\)

\(\Delta=\left(x+2\right)^2-4\left(3x-3\right)=\left(x-4\right)^2\)

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

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

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