7.1. Find the minimum value of the expression $\frac{5 x^{2}+8 x y+5 y^{2}-14 x-10 y+30}{\left(4-x^{2}-10 x y-25 y^{2}\right)^{7 / 2}}$. Round the answer to hundredths if necessary.
Answer: 0.16 (exact value: $\frac{5}{32}$).

Solution. The expression from the problem condition is equal to

$$
\frac{(2 x+y)^{2}+(x+2 y)^{2}-14 x-10 y+30}{\left(4-(x+5 y)^{2}\right)^{7 / 2}}=\frac{(2 x+y-3)^{2}+(x+2 y-1)^{2}+20}{\left(4-(x+5 y)^{2}\right)^{7 / 2}}
$$

The numerator of this fraction is no less than 20, and equality is achieved when $2 x+y-3=x+2 y-1=0$, i.e., when $x=\frac{5}{3}, y=-\frac{1}{3}$. For these same values of $x$ and $y$, the denominator reaches its maximum value $4^{7 / 2}=2^{7}=128$. Therefore, the minimum value of the fraction is $\frac{20}{128}=\frac{5}{32} \approx 0.16$.