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2 tháng 7 2015
A = \(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{9999}{10000}=\frac{1\cdot3}{2.2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{99\cdot101}{100\cdot100}=\frac{1}{2}\cdot\frac{101}{100}=\frac{101}{200}\)
B = ( 1- 1/4 )( 1-1/9) ...( 1-1/10000 ) = 3/4 . 8/9 .....9999/100000 ( tương tự A )
7 tháng 8 2020
\(a,\frac{-1}{9}.\frac{15}{22}.\frac{-9}{25}\)
\(=\frac{-1.15.\left(-9\right)}{9.22.25}\)
\(=\frac{3}{110}\)
\(b,\frac{-2}{7}.\left(\frac{5}{13}-\frac{9}{15}\right)-\frac{2}{7}.\frac{8}{13}\)
\(=\frac{-2}{7}.\left(\frac{5}{13}+\frac{8}{13}-\frac{3}{5}\right)\)
\(=\frac{-2}{7}.\left(1-\frac{3}{5}\right)\)
\(=\frac{-2}{7}.\frac{2}{5}\)
\(=\frac{-4}{35}\)
\(c,\frac{3}{10}.\left(\frac{-4}{9}+\frac{2}{5}\right)-\frac{3}{10}.\left(\frac{5}{9}-\frac{3}{5}\right)\)
\(=\frac{3}{10}.\left[\left(\frac{-4}{9}+\frac{2}{5}\right)-\left(\frac{5}{9}-\frac{3}{5}\right)\right]\)
\(=\frac{3}{10}.\left(\frac{-4}{9}+\frac{2}{5}-\frac{5}{9}+\frac{3}{5}\right)\)
\(=\frac{3}{10}.\left[\left(\frac{-4}{9}-\frac{5}{9}\right)+\left(\frac{2}{5}+\frac{3}{5}\right)\right]\)
\(=\frac{3}{10}.\left(-1+1\right)\)
\(=\frac{3}{10}.0\)
\(=0\)
\(d,\frac{4}{11}-\frac{5}{13}+\frac{7}{11}-\frac{8}{13}\)
\(=\left(\frac{4}{11}+\frac{7}{11}\right)+\left(\frac{-5}{13}-\frac{8}{13}\right)\)
\(=1-1\)
\(=0\)
Học tốt
LG
6 tháng 7 2017
1. \(\dfrac{9.5^{20}.27^9-3.9^{15}:25^9}{7.3^{29}.125^6-3.3^9.15^{19}}\)
\(=\dfrac{3^2.5^{20}.3^{27}-3.3^{30}.5^{18}}{7.3^{29}.5^{18}-3^{10}.3^{19}.5^{19}}\)
\(=\dfrac{3^{29}.5^{20}-3^{31}.5^{18}}{7.3^{29}.5^{18}-3^{29}.5^{19}}\)
\(=\dfrac{3^{28}.5^{18}.\left(5^2-3^2\right)}{3^{29}.5^{18}.\left(7-5\right)}\)
\(=\dfrac{5^2-3^2}{7-5}=\dfrac{16}{2}=8\)
2.\(\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{-4}\right)^{20}=\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}\)
\(=\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{2}^2\right)^{20}=\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{2}\right)^{40}\)
\(=\left(\dfrac{1}{2}\right)^{55}\)
3.\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)
\(=\dfrac{1-3}{1+5}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
Chúc học tốt!!
5 tháng 7 2017
\(\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^{2013}=\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)=-\dfrac{1}{6}\)
\(\left(\dfrac{1}{5}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}=\dfrac{1}{5^{12}}.\dfrac{1}{4^{20}}=5^{-12}.4^{-20}=125^{-4}.1024^{-4}=\left(125.1024\right)^{-4}=128000^{-4}\)
5 tháng 7 2017
\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.2^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\dfrac{2.6}{3.7}=\dfrac{4}{7}\)
7 tháng 3 2018
\(\frac{9.5^{20}.27^9-3.9^{15}.25^9}{7.3^{29}.125^6-3.3^9.15^{19}}\)
\(=\frac{3^2.5^{20}.3^{27}-3.3^{30}.5^{18}}{7.3^{29}.5^{18}-3^{10}.3^{19}.5^{19}}\)
\(=\frac{3^{29}.5^{18}.5^2-3^2.3^{29}.5^{18}}{7.3^{29}.5^{18}-3^{29}.5^{18}.5}\)
\(=\frac{3^{29}.5^{18}.\left(5^2-3^2\right)}{3^{29}.5^{18}.\left(7-5\right)}\)
\(=\frac{25-9}{2}\)
\(=\frac{16}{2}=8\)
b)\(\dfrac{1}{15}\)
c) \(\dfrac{3}{110}\)
b)\(=\dfrac{1}{15}\)
c)\(\dfrac{3}{110}\)