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\(=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{\left(-6\right)^{11}-2^{12}\cdot3^{12}}=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{\left(-6\right)^{11}-6^{12}}=\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{\left(-6\right)^{11}\left(1+6\right)}=\frac{6^{10}\cdot2^2\cdot2\cdot3}{\left(-6\right)^{11}\cdot7}=\frac{6^{11}\cdot4}{\left(-6\right)^{11}\cdot7}=\frac{-4}{7}\)
a) \(=\frac{\left(-2\right)^{10}}{\left(-2\right)^7}=\frac{\left(-2\right)^7.\left(-2\right)^3}{\left(-2\right)^7}=\left(-2\right)^3=-8\)
b) \(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2.3}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2.3}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{12}.3^{10}.\left(3^2-2^{-11}.3^{-9}\right)}=\frac{6}{3^2-2^{-11}.3^{-9}}\)
\(=\frac{2.3}{3.\left(3-2^{-11}.3^{-10}\right)}=\frac{2}{3-2^{-11}.3^{-10}}\)
\(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}^2-...-\frac{1}{5}\right)\left(2,4.42-21.4,8\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)
=> \(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-...-\frac{1}{5}\right).0}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)= 0
