Đặt \(\sqrt{x^2+9}=t>0\) ta được:
\(t^2+8x=\left(x+8\right)t\Leftrightarrow t^2-\left(x+8\right)t+8x=0\)
\(\Leftrightarrow t^2-tx-8t+8x=0\)
\(\Leftrightarrow t\left(t-x\right)-8\left(t-x\right)=0\)
\(\Leftrightarrow\left(t-x\right)\left(t-8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+9}=x\left(x\ge0\right)\\\sqrt{x^2+9}=8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+9=x^2\left(vn\right)\\x^2=55\end{matrix}\right.\)
\(\Rightarrow x=\pm\sqrt{55}\)
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