ĐKXĐ: \(-1\le x\le2\)
\(\Leftrightarrow2x^2-2x-1+\sqrt{\left(x+1\right)\left(2-x\right)}\le0\)
Đặt \(\sqrt{\left(x+1\right)\left(2-x\right)}=t\ge0\)
\(\Rightarrow2x^2-2x=4-2t^2\)
BPT trở thành:
\(4-2t^2-1+t\le0\Leftrightarrow-2t^2+t+3\le0\Rightarrow\left[{}\begin{matrix}t\le-1\left(l\right)\\t\ge\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)\left(2-x\right)\ge\frac{9}{4}\)
\(\Leftrightarrow x^2-x+\frac{1}{4}\le0\Rightarrow x=\frac{1}{2}\)
Vậy BPT có nghiệm duy nhất \(x=\frac{1}{2}\)
