Question

[:vvv]

Giải pt:

\(x^2+3x+1=\left(x+3\right)\sqrt{x^2+1}\)

Answer 1

[Nguyễn Việt Lâm]

Đặt \(\sqrt{x^2+1}=t>0\) ta được:

\(t^2+3x=\left(x+3\right)t\)

\(\Leftrightarrow t^2-\left(x+3\right)t+3x=0\)

\(\Delta=\left(x+3\right)^2-12x=\left(x-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{x+3-\left(x-3\right)}{2}=3\\t=\dfrac{x+3+x-3}{2}=x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=3\\\sqrt{x^2+1}=x\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=8\\x^2+1=x^2\left(vô-nghiệm\right)\end{matrix}\right.\)

\(\Rightarrow x=\pm2\sqrt{2}\)

Answer 2

[Viêt Thanh Nguyễn Hoàng]

\(pt\Leftrightarrow x^2+3x+1=x\sqrt{x^2+1}+3\sqrt{x^2+1}\Leftrightarrow\sqrt{x^2+1}\left(\sqrt{x^2+1}-x\right)-3\left(\sqrt{x^2+1}-x\right)=0\Leftrightarrow\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}-3\right)=0\)

Xét TH:

TH1 : \(\sqrt{x^2+1}-x=0\Leftrightarrow\sqrt{x^2+1}=x\Leftrightarrow x^2+1=x^2\left(x\ge0\right)\Leftrightarrow0=1\left(voli\right)\)

TH2 : \(\sqrt{x^2+1}-3=0\Leftrightarrow\sqrt{x^2+1}=3\Leftrightarrow x^2+1=9\Leftrightarrow x^2=8\Leftrightarrow x=\pm2\sqrt{2}\)