Đk: \(x\ge-1\)
\(x^2+\sqrt{x+1}=1\Leftrightarrow\left(x-1\right)\left(x+1\right)+\sqrt{x+1}=0\Leftrightarrow\left(x+1-2\right)\left(x+1\right)+\sqrt{x+1}=0\) (*)
Đặt \(t=\sqrt{x+1}\left(t\ge0\right)\)
pt (*) trở thành: \(\left(t^2-2\right)t^2+t=0\Leftrightarrow t\left(t^3-2t+1\right)=0\Leftrightarrow t\left(t-1\right)\left(t^2+t-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=1\\t=\dfrac{-1+\sqrt{5}}{2}\\t=\dfrac{-1-\sqrt{5}}{2}\left(L\right)\end{matrix}\right.\)
Với t=0, ta có: \(\sqrt{x+1}=0\Leftrightarrow x=-1\left(N\right)\)
Với t = 1, ta có: \(\sqrt{x+1}=1\Leftrightarrow x=0\left(N\right)\)
Với \(t=\dfrac{-1+\sqrt{5}}{2}\), ta có:
\(\sqrt{x+1}=\dfrac{-1+\sqrt{5}}{2}\Leftrightarrow x+1=\dfrac{3-\sqrt{5}}{2}\Leftrightarrow x=\dfrac{1-\sqrt{5}}{2}\left(N\right)\)
Kl: x=-1, x=0, \(x=\dfrac{1-\sqrt{5}}{2}\)
