ĐK: \(x\ge2\)
\(PT\Leftrightarrow\left(\sqrt{x+1}-\sqrt{x-2}\right)\left(1+\sqrt{x^2-x-2}\right)=\left(x+1\right)-\left(x-2\right)\)
\(\Leftrightarrow\left(\sqrt{x+1}-\sqrt{x-2}\right)\left(1+\sqrt{x^2-x-2}\right)=\left(\sqrt{x+1}-\sqrt{x-2}\right)\left(\sqrt{x+1}+\sqrt{x-2}\right)\)
Với \(\sqrt{x+1}=\sqrt{x-2}\) \(\Leftrightarrow x+1=x-2\) (loại)
Với \(\sqrt{x+1}-\sqrt{x-2}\ne0\) , chia cả 2 vế cho \(\sqrt{x+1}-\sqrt{x-2}\)
\(\Leftrightarrow1+\sqrt{x^2-x-2}=\sqrt{x+1}+\sqrt{x-2}\)
\(\Leftrightarrow x^2-x-1+2\sqrt{x^2-x-2}=2x-1+2\sqrt{x^2-x-2}\)
\(\)\(\Leftrightarrow x^2-3x=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=3\end{matrix}\right.\)
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