ĐKXĐ: \(-1\le x\le1\)
Đặt \(\sqrt{1-x^2}=a\ge0\) ta được:
\(\left\{{}\begin{matrix}x^2+a^2=1\\x^3+a^3=\sqrt{2}ax\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+a^2=1\\\left(x+a\right)\left(x^2+a^2-ax\right)=\sqrt{2}ax\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+a^2=1\\\left(x+a\right)\left(1-ax\right)=\sqrt{2}ax\end{matrix}\right.\)
Đặt \(x+a=t\Rightarrow x^2+a^2+2ax=t^2\Rightarrow ax=\frac{t^2-1}{2}\)
\(\Rightarrow t\left(1-\frac{t^2-1}{2}\right)=\sqrt{2}\left(\frac{t^2-1}{2}\right)\)
\(\Leftrightarrow t^3+\sqrt{2}t^2-3t-\sqrt{2}=0\)
\(\Leftrightarrow\left(t-\sqrt{2}\right)\left(t^2+2\sqrt{2}t+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=\sqrt{2}\\t=1-\sqrt{2}\\t=-1-\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+\sqrt{1-x^2}=\sqrt{2}\\x+\sqrt{1-x^2}=-1-\sqrt{2}\left(l\right)\\x+\sqrt{1-x^2}=1-\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1-x^2}=\sqrt{2}-x\\\sqrt{1-x^2}=1-\sqrt{2}-x\left(x\le1-\sqrt{2}\right)\\\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x^2=\left(\sqrt{2}-x\right)^2\\1-x^2=\left(1-\sqrt{2}-x\right)^2\end{matrix}\right.\) \(\Leftrightarrow...\)
