\(a,A=\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{7}-\dfrac{9}{11}+9}\)
\(=\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{9+\dfrac{9}{7}-\dfrac{9}{11}+\dfrac{9}{1001}-\dfrac{9}{13}}\)
\(=\dfrac{3\cdot\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}{9\cdot\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}\)
\(=\dfrac{3}{9}\)
\(=\dfrac{1}{3}\)
\(---\)
\(b,B=\dfrac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{7\cdot2^{29}\cdot3^{18}-5\cdot2^{28}\cdot3^{18}}\)
\(=\dfrac{5\cdot2^{18}\cdot3^{18}\cdot2^{12}-2\cdot2^{28}\cdot3^{14}\cdot3^4}{2^{28}\cdot3^{18}\cdot\left(7\cdot2-5\right)}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}\cdot\left(14-5\right)}\)
\(=\dfrac{2^{29}\cdot3^{18}\cdot\left(5\cdot2-1\right)}{2^{28}\cdot3^{18}\cdot9}\)
\(=\dfrac{2\cdot\left(10-1\right)}{9}\)
\(=\dfrac{2\cdot9}{9}\)
\(=2\)
\(---\)
\(c,C=\dfrac{5\cdot2^{30}\cdot3^{18}-4\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{2^{28}\cdot3^{18}\cdot\left(5\cdot3-7\cdot2\right)}\)
\(=\dfrac{2^{29}\cdot3^{18}\cdot\left(5\cdot2-3^2\right)}{2^{28}\cdot3^{18}\cdot\left(15-14\right)}\)
\(=\dfrac{2\cdot\left(10-9\right)}{1}\)
\(=2\)
\(---\)
\(d,D=\dfrac{15^{15}\cdot7^{16}}{6\cdot3^{14}\cdot35^{15}-15^8\cdot35^7\cdot7\cdot21^7}\)
\(=\dfrac{\left(3\cdot5\right)^{15}\cdot7^{16}}{2\cdot3\cdot3^{14}\cdot\left(5\cdot7\right)^{15}-\left(3\cdot5\right)^8\cdot\left(5\cdot7\right)^7\cdot7\cdot\left(3\cdot7\right)^7}\)
\(=\dfrac{3^{15}\cdot5^{15}\cdot7^{16}}{2\cdot3^{15}\cdot5^{15}\cdot7^{15}-3^8\cdot5^8\cdot5^7\cdot7^7\cdot7\cdot3^7\cdot7^7}\)
\(=\dfrac{3^{15}\cdot5^{15}\cdot7^{16}}{2\cdot3^{15}\cdot5^{15}\cdot7^{15}-3^{15}\cdot5^{15}\cdot7^{15}}\)
\(=\dfrac{3^{15}\cdot5^{15}\cdot7^{16}}{3^{15}\cdot5^{15}\cdot7^{15}\cdot\left(2-1\right)}\)
\(=\dfrac{7}{1}\)
\(=7\)
#\(Toru\)
