Question

\(\sqrt{-x^2-2x+3}\)≤x+3

Answer 1

nhầm 

 

Answer 2

ĐK: \(x\ge1;x\le-3\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+3\ge0\\-x^2-2x+3\le x^2+6x+9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-3\\x^2+4x+3\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-3\\\left[{}\begin{matrix}x\ge-1\\x\le-3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x\ge-1\end{matrix}\right.\)

Đối chiếu điều kiện ta được \(x=-3;x\ge1\)