Question

giải phương trình

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

Answer 1

Đặt \(a=\sqrt{1-x^2},b=x\Rightarrow a^2+b^2=1-x^2+x^2=1\).
Ta có \(\dfrac{1}{a}+\dfrac{1}{b}=2\Leftrightarrow\dfrac{a+b}{ab}=2\sqrt{2}\) \(\Leftrightarrow\dfrac{a^2+b^2+2ab}{a^2b^2}=8\) \(\Leftrightarrow\dfrac{1+2ab}{a^2b^2}=8\)
Suy ra \(8a^2b^2-2ab-1=0\) \(\Leftrightarrow\left[{}\begin{matrix}ab=-\dfrac{1}{4}\\ab=\dfrac{1}{2}\end{matrix}\right.\).
Với \(ab=-\dfrac{1}{4}\)ta có: \(\left\{{}\begin{matrix}ab=-\dfrac{1}{4}\\a^2+b^2=1\end{matrix}\right.\)
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Với \(ab=\dfrac{1}{2}\) ta có \(\left\{{}\begin{matrix}ab=\dfrac{1}{2}\\a^2+b^2=1\end{matrix}\right.\)