ĐKXĐ: ...
\(\Leftrightarrow x^2-4x-1=\sqrt{x+3}\)
Đặt \(\sqrt{x+3}=a-2\ge0\) ta được hệ:
\(\left\{{}\begin{matrix}x^2-4x-1=a-2\\x+3=\left(a-2\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-4x+1-a=0\\a^2-4a+1-x=0\end{matrix}\right.\)
Trừ vế cho vế:
\(\left(x-a\right)\left(x+a\right)-3\left(x-a\right)=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+a-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=a\\x+a-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+3}=x-2\\\sqrt{x+3}=1-x\end{matrix}\right.\)
\(\Leftrightarrow...\)
