Question

giải phương trình :

\(\sqrt{x+2}=3\)

Answer 1

\(\sqrt{x+2}=3\\ ĐK:x+2\ge0\Leftrightarrow x\ge-2\\ \sqrt{x+2}=3\\ \Leftrightarrow x+2=9\\ \Leftrightarrow x=9-2\\ \Leftrightarrow x=7\left(tm\right)\)

Vậy \(S=\left\{7\right\}\)

 

Answer 2

\(\sqrt{x+2}=3\) (ĐK: \(x\ge-2\))

\(\Leftrightarrow x+2=3^2\)

\(\Leftrightarrow x+2=9\)

\(\Leftrightarrow x=9-2\)

\(\Leftrightarrow x=7\left(tm\right)\)

Vậy \(x=7\)

Answer 3

\(\sqrt{x+2}=3\)

\(\RightarrowĐKXĐ:x\)\(\ge-2\)

\(\Leftrightarrow (\sqrt{x+2})^2=3^2\)

\(\Leftrightarrow x+2=9\)

\(\Leftrightarrow x=7\)

Vậy phương trình có nghiệm \(x=7\)

=((