\(ĐK:x\ge-1\)
Đặt \(a=\sqrt{x+1};b=\sqrt{x^2-x+1};c=\sqrt{x+3}\left(a,b,c\ge0\right)\)
\(PT\Leftrightarrow\dfrac{ab}{c}+c=a+b\Leftrightarrow ab+c^2=ac+bc\\ \Leftrightarrow ab+c^2-ac-bc=0\\ \Leftrightarrow\left(a-c\right)\left(b-c\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=c\\b=c\end{matrix}\right.\)
Với \(a=c\Leftrightarrow x+1=x+3\Leftrightarrow0x=2\Leftrightarrow x\in\varnothing\)
Với \(b=c\Leftrightarrow x^2-x+1=x+3\Leftrightarrow x^2-2x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1+\sqrt{3}\left(tm\right)\\x=1-\sqrt{3}\left(tm\right)\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{1\pm\sqrt{3}\right\}\)
