điều kiện xác định : \(x\ne0;\dfrac{1}{x^2}\ge\dfrac{3}{4}\)
ta có : \(\sqrt{\dfrac{1}{x^2}-\dfrac{3}{4}}=\dfrac{1}{x}-\dfrac{1}{2}\Leftrightarrow\dfrac{1}{x^2}-\dfrac{3}{4}=\left(\dfrac{1}{x}-\dfrac{1}{2}\right)^2\)
\(\Leftrightarrow\dfrac{1}{x^2}-\dfrac{3}{4}=\dfrac{1}{x^2}-\dfrac{1}{x}+\dfrac{1}{4}\Leftrightarrow\dfrac{1}{x}=1\Leftrightarrow x=1\left(tmđk\right)\)
vậy \(x=1\)