giải giúp mik

a) $\frac{2^4.9^3}{3^4.8^2}$ = $\frac{2^{4}.(3^2)^3}{(3^2)^2.(2^3)^2}$ = $\frac{2^{4}.3^{6}}{3^{4}.2^{6}}$ = $2^{4-6}.3^{6-4}$ = $2^{-2}.3^{2}$ = $\frac{1}{4}.9$ = $\frac{9}{4}$ b) $\frac{8^{24}.9^{28}}{27^{15}.4^{35}}$ = $\frac{(2^3)^{24}.(3^2)^{28}}{(3^3)^{15}.(2^2)^{35}}$ = $\frac{2^{72}.3^{56}}{3^{45}.2^{70}}$ = $2^{72-70}.3^{56-45}$ = $2^{2}.3^{11}$ = $4.177147$ c) $\frac{5^{72}.8^{34}}{(-2)^{100}.125^{25}}$ = $\frac{5^{72}.(2^3)^{34}}{(-2)^{100}.(5^3)^{25}}$ = $\frac{5^{72}.2^{102}}{-2^{100}.5^{75}}$ = $5^{72-75}.2^{102-100}$ = $5^{-3}.2^{2}$ = $\frac{1}{125}.4$ = $\frac{4}{125}$ d) $\frac{2^{12}.3^5-4^6.81}{(2^2.3)^6+8^4.3^5}$ = $\frac{2^{12}.3^5-(2^2)^6.(3^4)}{(2^2.3)^6+(2^3)^4.3^5}$ = $\frac{2^{12}.3^5-2^{12}.3^4}{(2^2.3)^6+2^{12}.3^5}$ = $\frac{2^{12}.3^5(1-3^{-1})}{2^{12}.3^5(1+2^{-1})}$ = $\frac{1-3^{-1}}{1+2^{-1}}$ = $\frac{2/3}{3/2}$ = $\frac{4}{9}$