Bài 5:
a: Ta có: \(\sqrt{x^2+6x+9}=3x-6\)
\(\Leftrightarrow\left|x+3\right|=3x-6\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-9\\4x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(nhận\right)\\x=\dfrac{3}{4}\left(loại\right)\end{matrix}\right.\)






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