ĐKXĐ: \(x\ge-3\)
\(x^2+2x\sqrt{x+3}+x+3+x+\sqrt{x+3}=12\)
\(\Leftrightarrow\left(x+\sqrt{x+3}\right)^2+x+\sqrt{x+3}-12=0\)
Đặt \(x+\sqrt{x+3}=t\ge-3\) ta được:
\(t^2+t-12=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-4\left(l\right)\end{matrix}\right.\) \(\Rightarrow x+\sqrt{x+3}=3\Leftrightarrow\sqrt{x+3}=3-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}3-x\ge0\\x+3=\left(3-x\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le3\\x^2-7x+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\left(l\right)\\x=1\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất x=1
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