\(x^2+\sqrt{x+2}=2\left(x\ge-2\right)\)
\(\Leftrightarrow\left(x^2-1\right)+\left(\sqrt{x+2}-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)+\dfrac{x+2-1}{\sqrt{x+2}-1}=0\)
\(\Leftrightarrow\left[\left(x-1\right)+\dfrac{1}{\sqrt{x+2}-1}\right]\left(x+1\right)=0\)
Pt \(\left(x-1\right)+\dfrac{1}{\sqrt{x+2}-1}=0\) vô no
=> x + 1 = 0
<=> x = - 1 (nhận)
\(x^2+\sqrt{x+2}=2\)
Đk:\(x\ge -2\)
\(pt\Leftrightarrow\sqrt{x+2}=2-x^2\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{2}\le x\le\sqrt{2}\\x+2=x^4-4x^2+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{2}\le x\le\sqrt{2}\\-x^4+4x^2+x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{2}\le x\le\sqrt{2}\\-\left(x-2\right)\left(x+1\right)\left(x^2+x-1\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-\sqrt{2}\le x\le\sqrt{2}\\\left[{}\begin{matrix}x=2\left(loai\right)\\x=-1\\x=\dfrac{\sqrt{5}-1}{2}\\x=-\dfrac{\sqrt{5}+1}{2}\left(loai\right)\end{matrix}\right.\end{matrix}\right.\)
