Question

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

Answer 1

\(\dfrac{9}{x^2}+\dfrac{2x}{\sqrt{2x^2+9}}-1=0\)

\(\Leftrightarrow\dfrac{9}{x^2}-2+\dfrac{2x}{\sqrt{2x^2+9}}+1=0\)

\(\Leftrightarrow\dfrac{-\left(2x^2-9\right)}{x^2}+\dfrac{\dfrac{2x^2-9}{2x^2+9}}{\dfrac{2x}{\sqrt{2x^2+9}}-1}=0\)

\(\Leftrightarrow\left(2x^2-9\right)\left(\dfrac{\dfrac{1}{2x^2+9}}{\dfrac{2x}{\sqrt{2x^2+9}}-1}-\dfrac{1}{x^2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x^2=9\\\dfrac{\dfrac{1}{2x^2+9}}{\dfrac{2x}{\sqrt{2x^2+9}-1}}=\dfrac{1}{x^2}\end{matrix}\right.\)\(\Rightarrow x=-\dfrac{3}{\sqrt{2}}\) (thỏa)