Question

Answer 1

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+1}=a>0\\\sqrt{x+3}=b\ge0\end{matrix}\right.\Leftrightarrow a^2+2b^2=x^2+2x+7\), PTTT:

\(a^2+2b^2=3ab\\ \Leftrightarrow a^2-3ab+2b^2=0\\ \Leftrightarrow a^2-ab-2ab+2b^2=0\\ \Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\)

Với \(a=b\Leftrightarrow x^2+1=x+3\Leftrightarrow x^2-x-2=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Với \(a=2b\Leftrightarrow x^2+1=4x+12\Leftrightarrow x^2-4x-11=0\Leftrightarrow x=2\pm\sqrt{15}\)

Vậy pt có nghiệm \(S=\left\{-1;2;2\pm\sqrt{15}\right\}\)