\(\sqrt{x+2}=3\\ ĐK:x+2\ge0\Leftrightarrow x\ge-2\\ \sqrt{x+2}=3\\ \Leftrightarrow x+2=9\\ \Leftrightarrow x=9-2\\ \Leftrightarrow x=7\left(tm\right)\)
Vậy \(S=\left\{7\right\}\)
\(\sqrt{x+2}=3\) (ĐK: \(x\ge-2\))
\(\Leftrightarrow x+2=3^2\)
\(\Leftrightarrow x+2=9\)
\(\Leftrightarrow x=9-2\)
\(\Leftrightarrow x=7\left(tm\right)\)
Vậy \(x=7\)
\(\sqrt{x+2}=3\)
\(\RightarrowĐKXĐ:x\)\(\ge-2\)
\(\Leftrightarrow (\sqrt{x+2})^2=3^2\)
\(\Leftrightarrow x+2=9\)
\(\Leftrightarrow x=7\)
Vậy phương trình có nghiệm \(x=7\)
=((
