\(PT\Leftrightarrow\left(x^2+2\right)\left(\sqrt{x^2+1}-x\right)=6x^2+11x+10\)
\(\Leftrightarrow\left(x^2+2\right)\sqrt{x^2+1}=x^3+6x^2+13x+10\)
\(\Leftrightarrow\left(\sqrt{x^2+1}\right)^3+\sqrt{x^2+1}=\left(x+2\right)^3+\left(x+2\right)\) (*)
Ta thấy nếu a > b thì \(a^3>b^3\Rightarrow a^3+a>b^3+b\)
Do đó \(\left(\cdot\right)\Leftrightarrow\sqrt{x^2+1}=x+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x^2+1=x^2+4x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x=-\dfrac{3}{4}\left(TM\right)\end{matrix}\right.\)
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