Solve the following equation in the set of real numbers:

$$
\frac{1}{x^{2}}+\frac{1}{x^{6}}+\frac{1}{x^{8}}+\frac{1}{x^{12}}-\frac{4}{x^{7}}=0
$$
Based on the arithmetic and geometric mean inequality,

$$
\frac{1}{x^{2}}+\frac{1}{x^{6}}+\frac{1}{x^{8}}+\frac{1}{x^{12}} \geq 4 \sqrt[4]{\frac{1}{x^{2}} \cdot \frac{1}{x^{6}} \cdot \frac{1}{x^{8}} \cdot \frac{1}{x^{12}}}=\frac{4}{|x|^{7}} \geq \frac{4}{x^{7}}
$$

thus, the left side is always a non-negative number.

Equality holds if the four numbers in the means are equal: $\frac{1}{x^{2}}=\frac{1}{x^{6}}=\frac{1}{x^{8}}=\frac{1}{x^{12}}$, which means $|x|=1$, and $x$ is positive. Therefore, the only solution is $x=1$.