3.已知$3x+5y-1=0$，求$8^x · 32^y$的值.
答案：解:$8^x · 32^y = 2^{3x} · 2^{5y} = 2^{3x+5y}$,
$\because 3x+5y-1=0,\therefore 3x+5y=1,\therefore 原式=2^1=2.$
4.计算：
(1)$5^9 × 0.2^8$；
(2)$(\dfrac{3}{7})^{2024}$
$(-2\dfrac{1}{3})^{2025}$。
答案：解:(1)原式$=(5×0.2)^8 ×5=1×5=5.$
(2)原式$=(\dfrac{3}{7})^{2024} × (-\dfrac{7}{3})^{2025}$
$=-(\dfrac{3}{7})^{2024} × (\dfrac{7}{3})^{2025}$
$=-(\dfrac{3}{7})^{2024} × (\dfrac{7}{3})^{2024} × \dfrac{7}{3}$
$=-(\dfrac{3}{7} × \dfrac{7}{3})^{2024} × \dfrac{7}{3}$
$=-\dfrac{7}{3}.$
5.若$a^{2n}=3$，$b^n=\frac{1}{4}$，求$(-ab)^{2n}$的值.
答案：解:$\because a^{2n}=3,b^n=\dfrac{1}{4},\therefore (-ab)^{2n}=(ab)^{2n}=a^{2n} · b^{2n}=$
$a^{2n} · (b^n)^2=3×(\dfrac{1}{4})^2=3×\dfrac{1}{16}=\dfrac{3}{16}.$
