Question

Tìm ĐKXĐ: \(\sqrt{x+2}=\sqrt{6-x}\)

Answer 1

x ∈ [ -2; 6 ]

Answer 2

ĐKXĐ: \(\left\{{}\begin{matrix}x+2\ge0\\6-x\ge0\end{matrix}\right.\)\(\Leftrightarrow6\ge x\ge-2\)

Answer 3

Answer 4

ĐKXĐ: \(-2\le x\le6\)

Answer 5

ĐK: \(\left\{{}\begin{matrix}x+2\ge0\\6-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x\le6\end{matrix}\right.\Leftrightarrow-2\le x\le6\)