ĐK:\(x\ge-1\)
\(x^2+\sqrt{x+1}=1\Leftrightarrow x^2+2x+1=2x+2-\sqrt{x+1}\Leftrightarrow\left(x+1\right)^2=2\left(x+1\right)-\sqrt{x+1}\left(1\right)\)
Đặt \(t=\sqrt{x+1}\left(t\ge0\right)\)
Vậy (1)\(\Leftrightarrow t^4=2t^2-t\Leftrightarrow t^4-2t^2+t=0\Leftrightarrow t \left(t^3-2t+1\right)=0\Leftrightarrow t\left(t-1\right)\left(t^2+t-1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}t=0\\t-1=0\\t^2+t-1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}t=0\left(tm\right)\\t=1\left(tm\right)\\t=\frac{\sqrt{5}-1}{2}\left(tm\right)\\t=\frac{-1-\sqrt{5}}{2}\left(ktm\right)\end{matrix}\right.\)
* t=0\(\Leftrightarrow\sqrt{x+1}=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\left(tm\right)\)
* t=1\(\Leftrightarrow\sqrt{x+1}=1\Leftrightarrow x+1=1\Leftrightarrow x=0\left(tm\right)\)
* \(t=\frac{\sqrt{5}-1}{2}\Leftrightarrow\sqrt{x+1}=\frac{\sqrt{5}-1}{2}\Leftrightarrow x+1=\frac{3-\sqrt{5}}{2}\Leftrightarrow x=\frac{1-\sqrt{5}}{2}\left(tm\right)\)Vậy S={-1;\(\frac{1-\sqrt{5}}{2}\);0}
ĐK : \(x\ge-1\)
\(PT\Leftrightarrow\sqrt{x+1}=1-x^2\) ( ĐK : \(-1\le x\le1\)
\(\Leftrightarrow x^4-2x^2+1=x+1\)
\(\Leftrightarrow x^4-2x^2-x=0\)
\(\Leftrightarrow x\left(x^3-2x-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x-1\right)=0\)
Dễ thấy \(x^2-x-1\ne0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
