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解：​$(-a^{m+1})^2 ÷ [(a^m)^2\ \mathrm {·} a] $​
​$= a^{2m+2} ÷ (a^{2m}\ \mathrm {·} a) $​
​$= a^{2m+2} ÷ a^{2m+1} $​
​$= a$​
解：​$\frac {2^{20} × 0.25^{12}}{0.5^{11} × 4^3} $​
​$= \frac {2^{20} × (2^{-2})^{12}}{(2^{-1})^{11} × (2^2)^3} $​
​$= \frac {2^{20} × 2^{-24}}{2^{-11} × 2^6}$​
​$ = \frac {2^{-4}}{2^{-5}} $​
​$= 2$​
解：$∵2^m × 32 × 4^m = 2^m × 2^5 × (2^2)^m = 2^m × 2^5 × 2^{2m} = 2^{3m+5}$
又$∵2^m × 32 × 4^m = 2^{20}$
$∴3m+5=20，$
解得$m=5$
则$(-m^3)^2 ÷ (-m)^3 = m^6 ÷ (-m^3) = -m^3 = -5^3 = -125$
解：$∵10^m = 20，$$10^n = \frac{1}{5}$
$∴10^m ÷ 10^n = 20 ÷ \frac{1}{5} = 100，$即$10^{m-n}=10^2$
$∴m-n=2，$则$2n-2m=-4$
$∴9^n ÷ 3^{2m} = (3^2)^n ÷ 3^{2m} = 3^{2n} ÷ 3^{2m} = 3^{2n-2m} = 3^{-4} = \frac{1}{81}$
解：∵​$x=2^{m+1}，$​
∴​$2^m = \frac {x}{2}$​
又∵​$y=3+4^m，$​​$4^m=(2^m)^2$​
∴​$y=3+(\frac {x}{2})^2 = \frac {1}{4}x^2 + 3$​
解：$∵3^{3x+1} × 5^{3x+1} = (3 × 5)^{3x+1} = 15^{3x+1}$
又$∵3^{3x+1} × 5^{3x+1} = 15^{2x+4}$
$∴15^{3x+1}=15^{2x+4}$
$∴3x+1=2x+4，$解得$x=3$
解：$∵2^a × 27^b × 37^c × 47^d = 2^a × 3^{3b} × 37^c × 47^d，$且$1998=2 × 3^3 × 37$
$∴2^a × 3^{3b} × 37^c × 47^d = 2 × 3^3 × 37$
$∴a=1，$$b=1，$$c=1，$$d=0$
则$(a-b-c+d)^{203}=(1-1-1+0)^{203}=(-1)^{203}=-1$
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