Đặt \(\sqrt{x^2+3x+5}=t>0\Rightarrow2x^2+6x=2t^2-10\)
Pt trở thành:
\(2t^2-10-11t+25=0\Leftrightarrow2t^2-11t+15=0\)
\(\Rightarrow\left[{}\begin{matrix}t=3\\t=\frac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+3x+5}=3\\\sqrt{x^2+3x+5}=\frac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3x-4=0\\x^2+3x-\frac{5}{4}=0\end{matrix}\right.\) (bấm máy)
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