ĐKXĐ: ...
Đặt \(\sqrt{x+1}+\sqrt{3-x}=a>0\)
\(\Rightarrow a^2=4+2\sqrt{\left(x+1\right)\left(3-x\right)}\Rightarrow\sqrt{\left(x+1\right)\left(x-3\right)}=\frac{a^2-4}{2}\)
Phương trình trở thành:
\(a-\frac{a^2-4}{2}=1\Rightarrow a^2-2a-2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=1+\sqrt{3}\\a=1-\sqrt{3}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+1}+\sqrt{3-x}=\sqrt{3}+1\)
\(\Leftrightarrow4+2\sqrt{-x^2+2x+3}=4+2\sqrt{3}\)
\(\Leftrightarrow\sqrt{-x^2+2x+3}=\sqrt{3}\)
\(\Leftrightarrow-x^2+2x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
