ĐK: \(-2< x< 2\)
Đặt \(c=\sqrt{2-x^2}\left(c>0\right)\)
\(\Leftrightarrow c^2=2-x^2\Rightarrow x=\pm\sqrt{2-c^2}\)
Ta có: \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{c}=2\\x^2+c^2=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{c+x}{cx}=2\\x^2+c^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}c+x=2cx\\\left(x+c\right)^2=2+2cx\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}c+x=2cx\\4c^2x^2-2cx-2=0\end{matrix}\right.\)
\(\Rightarrow cx=1;cx=-\dfrac{1}{2}\)
+) \(cx=1\Rightarrow c=\dfrac{1}{x}\left(x\ne0\right)\)
\(\Leftrightarrow\dfrac{1}{x}+x=2\Leftrightarrow x^2-2x+1=0\Leftrightarrow x=1\)
+) \(cx=-\dfrac{1}{2}\Rightarrow x=-\dfrac{1}{2x}\left(x\ne0\right)\)
\(\Leftrightarrow\dfrac{1}{x}-2x=2\Leftrightarrow2x^2+2x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{-1+\sqrt{3}}{2}\rightarrow c< 0\left(loai\right)\\x_2=\dfrac{-1-\sqrt{3}}{2}\left(tm\right)\end{matrix}\right.\)
Vậy pt có 2 nghiệm....
