ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{1-x}=a\ge0\\\sqrt{1+x}=b\ge0\end{matrix}\right.\) \(\Rightarrow-x^2=a^2b^2-1\)
Ta được hệ:
\(\left\{{}\begin{matrix}a^2+b^2=2\\a+b=2+\frac{a^2b^2-1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-2ab=2\\4\left(a+b\right)-a^2b^2=7\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a+b=u\\ab=v\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}u^2-2v=2\\4u-v^2=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}v=\frac{u^2-2}{2}\\4u-v^2=7\end{matrix}\right.\)
\(\Leftrightarrow4u-\left(\frac{u^2-2}{2}\right)=7\)
\(\Leftrightarrow4u-\frac{u^4-4u^2+4}{4}=7\)
\(\Leftrightarrow u^4-4u^2-16u+32=0\)
\(\Leftrightarrow\left(u-2\right)\left(u^3+2u^2\right)-16\left(u-2\right)=0\)
\(\Leftrightarrow\left(u-2\right)\left(u^3+2u^2-16\right)=0\)
\(\Leftrightarrow\left(u-2\right)^2\left(u^2+4u+8\right)=0\Rightarrow u=2\Rightarrow v=1\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2\\ab=1\end{matrix}\right.\) \(\Rightarrow a=b=1\Rightarrow\sqrt{1-x}=1\Rightarrow x=0\)
