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\(=\dfrac{5\cdot2^{30}-3^{20}\cdot2^{29}}{5\cdot2^{10}-7\cdot2^{30}\cdot3^3}\)
\(=\dfrac{2^{29}\left(5\cdot2-3^{20}\right)}{2^{10}\left(5\cdot1-7\cdot2^{20}\cdot3^3\right)}=\dfrac{2^{19}\cdot\left(10-3^{20}\right)}{5-189\cdot2^{20}}\)
Rút gọn các biểu thức sau:
E = \(\dfrac{9.5^{20}.27^9-3.9^{15}.25^9}{7.3^{29}.125^6-3.3^9.15^{19}}\)
\(=\dfrac{3^2\cdot5^{20}\cdot3^{27}-3\cdot3^{30}\cdot5^{18}}{7\cdot3^{29}\cdot5^{13}-3^{29}\cdot5^{19}}=\dfrac{3^{29}\cdot5^{20}-5^{18}\cdot3^{31}}{3^{29}\cdot5^{13}\cdot7-3^{29}\cdot5^{19}}\)
\(=\dfrac{3^{29}\cdot5^{18}\left(5^2-3^2\right)}{3^{29}\cdot5^{13}\left(7-5^6\right)}=5^5\cdot\dfrac{4^2}{7-5^6}\)
a: \(=\left(\dfrac{1}{15}+\dfrac{14}{15}\right)+\left(\dfrac{9}{10}-2-\dfrac{11}{9}\right)+\dfrac{1}{157}\)
\(=1+\dfrac{1}{157}+\dfrac{81-180-110}{90}\)
\(=\dfrac{158}{157}+\dfrac{-209}{90}\simeq-1.315\)
b: \(=\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{2}{6}\)
=1/3-1/3
=0
c: \(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2015\cdot2017}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
=2016/2017
\(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\\ =\dfrac{2\left(0,2-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(0,2-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{2\left(\dfrac{1}{6}-0,125+0,1\right)}{7\left(\dfrac{1}{6}-0,125+0,1\right)}\\ =\dfrac{2}{7}-\dfrac{2}{7}\\ =0\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-3^{11}\cdot2^{11}}=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot5}=\dfrac{2}{3}\cdot\dfrac{6}{5}=\dfrac{4}{5}\)
\(=\frac{5\cdot2^{20}\cdot3^{11}+7\cdot2^{29}\cdot3^{18}}{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}=\frac{2^{20}\cdot3^{11}\left(5+7\cdot2^9\cdot3^7\right)}{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}=\frac{5+7\cdot2^9\cdot3^7}{2^9\cdot3^7\cdot1}\)