7. Let $f(x)$ satisfy the equation $f(x)-2 f\left(\frac{1}{x}\right)=x$. Then the range of $f(x)$ is $\qquad$ .
$=7 .\left(-\infty,-\frac{2 \sqrt{2}}{3}\right] \cup\left[\frac{2 \sqrt{2}}{3},+\infty\right)$.
From $f(x)-2 f\left(\frac{1}{x}\right)=x$, by replacing $x$ with $\frac{1}{x}$ we have
$$
f\left(\frac{1}{x}\right)-2 f(x)=\frac{1}{x} \text {. }
$$

Eliminating $f\left(\frac{1}{x}\right)$ from the two equations, we get $f(x)=-\frac{1}{3}\left(x+\frac{2}{x}\right)$. Therefore, $|f(x)|=\frac{1}{3}\left|x+\frac{2}{x}\right| \geqslant \frac{2 \sqrt{2}}{3}$.