\(x^2+x-\sqrt{x^2+x+1}-1=0\)
\(\Leftrightarrow x^2+x+1-2=\sqrt{x^2+x+1}\)(*)
Đặt \(\sqrt{x^2+x+1}=a\left(a\ge0\right)\)
(*)\(\Leftrightarrow a^2-2=a\)
\(\Leftrightarrow a^2-a-2=0\)
\(\Leftrightarrow\left(a-2\right)\left(a+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\left(TM\right)\\a=-1\left(L\right)\end{matrix}\right.\)
Trả lại biến cũ: \(\sqrt{x^2+x+1}=2\)\(\Leftrightarrow x^2+x+1=4\Leftrightarrow x^2+x-3=0\Leftrightarrow x=\frac{-1\pm\sqrt{13}}{2}\)