Question

Giải bất phương trình sau: \(\sqrt{8+2x-x^2}\le6-3x\)

Answer 1

\(\sqrt{8+2x-x^2}\le6-3x\)

⇒ \(\left\{{}\begin{matrix}-x^2+2x+8\ge0\\6-3x\ge0\\-x^2+2x+8\le\left(6-3x\right)^2\end{matrix}\right.\)

⇌ \(\left\{{}\begin{matrix}-2\le x\le4\\x\le2\\-x^2+2x+8\le36-36x+9x^2\end{matrix}\right.\)

⇌ \(\left\{{}\begin{matrix}-2\le x\le4\\x\le2\\-10x^2+38x-28\le0\end{matrix}\right.\)

⇌ \(\left\{{}\begin{matrix}-2< x< 4\\x\le2\\\left[{}\begin{matrix}x\le1\\x\ge\frac{14}{5}\end{matrix}\right.\end{matrix}\right.\)

⇌ \(-2\le x\le1\)

Vậy \(S=\left[-2;1\right]\)