Question

Giải phương trình: \(\frac{9}{x^2}+\frac{2x}{\sqrt{2x^2+9}}-1=0\)

Answer 1

ĐKXĐ: \(x\ne0\)

\(\Leftrightarrow\frac{2x^2+9}{x^2}+\frac{2x}{\sqrt{2x^2+9}}-3=0\)

Đặt \(\frac{x}{\sqrt{2x^2+9}}=a\Rightarrow\frac{2x^2+9}{x^2}=\frac{1}{a^2}\)

\(\frac{1}{a^2}+2a-3=0\)

\(\Leftrightarrow2a^3-3a^2+1=0\)

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

\(\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2x^2+9}\left(x>0\right)\\-2x=\sqrt{2x^2+9}\left(x< 0\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=2x^2+9\left(vn\right)\\2x^2=9\end{matrix}\right.\) \(\Rightarrow x=\frac{-3\sqrt{2}}{2}\)