\(x^2+\sqrt{x+1}=1\)
Đk:\(x\ge-1\)
\(\Leftrightarrow\sqrt{x+1}=1-x^2\left(-1\le x\le1\right)\)
\(\Leftrightarrow x+1=x^4-2x^2+1\)
\(\Leftrightarrow-x^4+2x^2+x=0\)
\(\Leftrightarrow-x\left(x^3-2x-1\right)=0\)
\(\Leftrightarrow-x\left(x+1\right)\left(x^2-x-1\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x=-1\\x=-\frac{\sqrt{5}-1}{2}\\x=0\end{matrix}\right.\)(thỏa mãn)