9. （1）已知$2^m\times32\times4^m = 2^{20}$，求$(-m^3)^2\div(-m)^3$的值；
（2）若$10^m = 20$，$10^n=\frac{1}{5}$，求$9^n\div3^{2m}$的值；
（3）已知$x = 2^{m + 1}$，$y = 3 + 4^m$，用含$x$的代数式表示$y$.
（2）若$10^m = 20$，$10^n=\frac{1}{5}$，求$9^n\div3^{2m}$的值；
（3）已知$x = 2^{m + 1}$，$y = 3 + 4^m$，用含$x$的代数式表示$y$.
答案
(1) ∵ $2^{m}×32×4^{m}=2^{3m + 5}=2^{20}$，∴ $3m + 5 = 20$，解得 $m = 5$，
∴ $(-m^{3})^{2}÷(-m)^{3}=m^{6}÷(-m)^{3}=-m^{3}=-5^{3}=-125$
(2) 由 $10^{m}=20$，$10^{n}=\frac{1}{5}$，得 $10^{m}÷10^{n}=20÷\frac{1}{5}$，即 $10^{m - n}=10^{2}$，∴ $m - n = 2$，∴ $2m - 2n = 4$，∴ $2n - 2m = -4$，∴ $9^{n}÷3^{2m}=3^{2n}÷3^{2m}=3^{2n - 2m}=3^{-4}=\frac{1}{81}$ (3) 由 $x = 2^{m + 1}$，$y = 3 + 4^{m}$，得 $2^{m}=\frac{x}{2}$，$y = 3+(2^{m})^{2}$，∴ $y = 3+(\frac{x}{2})^{2}=\frac{1}{4}x^{2}+3$
∴ $(-m^{3})^{2}÷(-m)^{3}=m^{6}÷(-m)^{3}=-m^{3}=-5^{3}=-125$
(2) 由 $10^{m}=20$，$10^{n}=\frac{1}{5}$，得 $10^{m}÷10^{n}=20÷\frac{1}{5}$，即 $10^{m - n}=10^{2}$，∴ $m - n = 2$，∴ $2m - 2n = 4$，∴ $2n - 2m = -4$，∴ $9^{n}÷3^{2m}=3^{2n}÷3^{2m}=3^{2n - 2m}=3^{-4}=\frac{1}{81}$ (3) 由 $x = 2^{m + 1}$，$y = 3 + 4^{m}$，得 $2^{m}=\frac{x}{2}$，$y = 3+(2^{m})^{2}$，∴ $y = 3+(\frac{x}{2})^{2}=\frac{1}{4}x^{2}+3$
10. 求等式中$x$的值：$3^{3x + 1}\times5^{3x + 1}=15^{2x + 4}$.
答案
∵ $3^{3x + 1}×5^{3x + 1}=15^{3x + 1}$，$3^{3x + 1}×5^{3x + 1}=15^{2x + 4}$，∴ $15^{3x + 1}=15^{2x + 4}$，∴ $3x + 1 = 2x + 4$，解得 $x = 3$
