\(A=\dfrac{12^{15}\cdot3^4-4^5\cdot3^9}{27^3\cdot2^{10}-32^3\cdot3^9}\\ =\dfrac{\left(2^2\cdot3\right)^{15}\cdot3^4-\left(2^2\right)^5\cdot3^9}{\left(3^3\right)^3\cdot2^{10}-\left(2^5\right)^3\cdot3^9}\\ =\dfrac{2^{30}\cdot3^{15}\cdot3^4-2^{10}\cdot3^9}{3^9\cdot2^{10}-2^{15}\cdot3^9}\\ =\dfrac{3^9\cdot2^{10}\left(2^{20}\cdot3^{10}\right)}{3^9\cdot2^{10}\left(1-2^5\right)}\\ =\dfrac{\left(2^2\right)^{10}\cdot3^{10}}{1-32}\\ =\dfrac{\left(2^2\cdot3\right)^{10}}{-31}\\ =\dfrac{-12^{10}}{31}\)
\(B=\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{99}{49^2\cdot50^2}\\ =\dfrac{2^2-1^2}{1^2\cdot2^2}+\dfrac{3^2-2^2}{2^2\cdot3^2}+...+\dfrac{50^2-49^2}{49^2\cdot50^2}\\ =\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{49^2}-\dfrac{1}{50^2}\\ =1-\dfrac{1}{2500}\\ =\dfrac{2499}{2500}\)
