Question

giải phương trình

\(\dfrac{1}{x}+\dfrac{1}{\sqrt{2-x^2}}=2\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}\left|x\right|< 2\\\dfrac{1}{\sqrt{2-x^2}}=2-\dfrac{1}{x}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\in\left(-2;]U[\dfrac{1}{2};2\right)\\\dfrac{1}{2-x^2}=\dfrac{\left(2x-1\right)^2}{x^2}\end{matrix}\right.\)

\(\Leftrightarrow x^2=\left(2-x^2\right)\left(4x^2-4x+1\right)\)

\(\Leftrightarrow\left(x-1\right)^2\left(2x^2+2x-1\right)=0\)

x=1 nhận

\(\left[{}\begin{matrix}x=1\\x=\dfrac{-1-\sqrt{3}}{2}\end{matrix}\right.\)