Question

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

Answer 1

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

Đặt \(\sqrt{x^2+1}=t>0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2\left(vn\right)\\x^2+1=9\end{matrix}\right.\) \(\Rightarrow x=...\)