ĐKXĐ: \(x\ne0\)
\(\dfrac{1}{x^2}+\dfrac{1}{x^2+2}=\dfrac{1}{12}\) (1)
\(\Leftrightarrow\dfrac{1}{x^2}+\dfrac{1}{x^2+2}-\dfrac{1}{12}=0\)
\(\Leftrightarrow\dfrac{12\left(x^2+2\right)+12x^2-x^2\left(x^2+2\right)}{12x^2\left(x^2+2\right)}=0\)
\(\Leftrightarrow\dfrac{12x^2+24+12x^2-x^4-2x^2}{12x^2\left(x^2+2\right)}=0\)
\(\Leftrightarrow\dfrac{22x^2+24-x^4}{12x^2\left(x^2+2\right)}=0\)
\(\Leftrightarrow-x^4+22x^2+24=0\)
đặt t = x2
\(\Leftrightarrow-t^2+22t+24=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=11+\sqrt{145}\\t=11-\sqrt{145}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=11+\sqrt{145}\\x^2=11-\sqrt{145}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{11+\sqrt{145}}\\x=-\sqrt{11+\sqrt{145}}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\sqrt{11+\sqrt{145}};-\sqrt{11+\sqrt{145}}\right\}\)
