2.109. $\left(-4 a^{3} \sqrt{\frac{\sqrt{a x}}{a^{2}}}\right)^{3}+\left(-10 a \sqrt{x} \cdot \sqrt{(a x)^{-1}}\right)^{2}+\left(-2\left(\sqrt[3]{a^{4} \sqrt{\frac{x}{a}}}\right)^{2}\right)^{3} ;$

$$
a=3 \frac{4}{7} ; x=0.28
$$
## Solution.

$$
\begin{aligned}
& \left(-4 a^{3} \sqrt{\frac{\sqrt{a x}}{a^{2}}}\right)^{3}+\left(-10 a \sqrt{x} \cdot \sqrt{(a x)^{-1}}\right)^{2}+\left(-2\left(\sqrt[3]{a \sqrt{\frac{x}{a}}}\right)^{2}\right)^{3}= \\
& =\frac{-64 a^{3} \sqrt{a x}}{a^{2}}+\frac{100 a^{2} x}{a x}-\frac{8 a^{2} \sqrt{x}}{\sqrt{a}}=-64 a \sqrt{a x}+100 a-8 a \sqrt{a x}= \\
& =100 a-72 a \sqrt{a x}=100 \cdot 3 \frac{4}{7}-72 \cdot 3 \frac{4}{7} \cdot \sqrt{3 \frac{4}{7} \cdot 0.28}= \\
& =100 \cdot \frac{25}{7}-72 \cdot \frac{25}{7} \cdot \sqrt{\frac{25}{7} \cdot \frac{7}{25}}=\frac{2500}{7}-\frac{1800}{7}=\frac{700}{7}=100
\end{aligned}
$$

Answer: 100.