10.7. Solve in $\square$ the equation $16 x+(x-1) \cdot 4^{x+1}=x^{2}\left(4^{x}+8+4^{\frac{1}{x}}\right)$.
Solution: DVA: $x \neq 0$.

The equation can be written in the form $x^{2} \cdot 4^{\frac{1}{x}}+(2-x)^{2} \cdot 4^{x}=8 x(2-x)$.

We observe that $x=2$ is not a solution.

Since $x \neq 2$, the equation can be brought to the form $\frac{x}{2-x} \cdot 4^{\frac{1}{x}}+\frac{2-x}{x} \cdot 4^{x}=8$.

If $\frac{x}{2-x} > 0$, that is, $x \in(0 ; 2)$. Using the relationship between the arithmetic mean and the geometric mean, we get $\frac{x}{2-x} \cdot 4^{\frac{1}{x}}+\frac{2-x}{x} \cdot 4^{x} \geq 2 \cdot \sqrt{\frac{x}{2-x} \cdot 4^{\frac{1}{x}} \cdot \frac{2-x}{x} \cdot 4^{x}}=2 \cdot \sqrt{4^{\frac{1}{x}+x}}=2^{\frac{1}{x}+x+1}$.

Since $x \in(0 ; 2)$, it follows that $x+\frac{1}{x} \geq 2$. Therefore, $\frac{x}{2-x} \cdot 4^{\frac{1}{x}}+\frac{2-x}{x} \cdot 4^{x} \geq 2^{2+1}=8$.

Equality holds when $\left\{\begin{array}{rl}\frac{x}{2-x} \cdot 4^{\frac{1}{x}} & =\frac{2-x}{x} \cdot 4^{x} \\ x & =1\end{array} \Leftrightarrow x=1 \in(0 ; 2)\right.$.

Answer: $S=\{1\}$.