Question

Câu 11 : A

Answer 1

\(\sqrt{3+2x-x^2}\\ =>3+2x-x^2\ge0\\ =>-\left(x^2-2x-3\right)\ge0\\ =>x^2-2x-3\le0\\ =>x^2-3x+x-3\le0\\ =>\left(x+1\right)\left(x-3\right)\le0\\ =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\ge0\\x-3\le0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\le0\\x-3\ge0\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-1\\x\le3\end{matrix}\right.\left(thoaman\right)\\\left\{{}\begin{matrix}x\le-1\\x\ge3\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

\(\\ =>B\)

Answer 2

Chọn C em nhé!