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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
H
1
11 tháng 3 2022
\(M=\dfrac{5^4\cdot50}{5^3\cdot15}=\dfrac{50}{3}>\dfrac{50}{4}=N\)
NN
0
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!!
BD
1
7 tháng 4 2018
chắc thế này :
\(A=1\frac{13}{15}\cdot\frac{1}{4}\cdot\left|-3\right|-\left(\frac{8}{15}-25\%\right)\div\frac{17}{40}\)
\(A=\frac{28}{15}\cdot\frac{1}{4}\cdot3-\left(\frac{8}{15}-\frac{25}{100}\right)\div\frac{17}{40}\)
\(A=\frac{7}{15}\cdot3-\left(\frac{8}{15}-\frac{1}{4}\right)\div\frac{17}{40}\)
\(A=\frac{21}{15}-\frac{17}{60}\div\frac{17}{40}\)
\(A=\frac{21}{15}-\frac{17}{60}\cdot\frac{40}{17}\)
\(A=\frac{21}{15}-\frac{40}{60}\)
\(A=\frac{21}{15}-\frac{2}{3}\)
\(A=\frac{11}{15}\)
TT
1
S
14 tháng 3 2017
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)
\(=\frac{1.2.3....99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4...100}\)
\(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).....\left(1-\frac{1}{10000}\right)\)
\(=\frac{3}{4}.\frac{8}{9}....\frac{9999}{10000}=\frac{101}{200}\)
OY
24 tháng 7 2021
A=\(\dfrac{3}{4}.\dfrac{8}{9}.....\dfrac{9999}{10000}\)
A=\(\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.....\dfrac{99.101}{100.100}\)
A=\(\dfrac{1.2.3.....99}{2.3.4.....100}.\dfrac{3.4.....101}{2.3.4.....100}\)
A=\(\dfrac{1}{100}.\dfrac{101}{2}\)
A=\(\dfrac{101}{200}\)
24 tháng 7 2021
\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{99.101}{100.100}\\ =\dfrac{1}{2}.\dfrac{101}{100}=\dfrac{101}{200}\)
\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)...\left(1-\dfrac{1}{10000}\right)\\ =\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
(làm như câu a)
b)\(\dfrac{1}{15}\)
c) \(\dfrac{3}{110}\)
b)\(=\dfrac{1}{15}\)
c)\(\dfrac{3}{110}\)