ĐK:\(x\ge\sqrt{8}-1\)
\(\sqrt{x^2+2x-7}+x^2+x-6\le\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{x^2+2x-7}-\sqrt{x-1}+x^2+x-6\le0\)
\(\Leftrightarrow\dfrac{x^2+x-6}{\sqrt{x^2+2x-7}+\sqrt{x-1}}+x^2+x-6\le0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(\dfrac{1}{\sqrt{x^2+2x-7}+\sqrt{x-1}}+1\right)\le0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(\dfrac{1}{\sqrt{x^2+2x-7}+\sqrt{x-1}}+1\right)\le0\left(1\right)\)
Dễ thấy: \(\dfrac{1}{\sqrt{x^2+2x-7}+\sqrt{x-1}}+1>0\forall x\forall\sqrt{8}-1\)
\(\Rightarrow\left(1\right)\Leftrightarrow\left(x-2\right)\left(x+3\right)\le0\)
Mà \(x+3>x-2\forall x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3\ge0\\x-2\le0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-3\\x\le2\end{matrix}\right.\)
