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21 tháng 8 2020

\(64^2.81^3.34\div2^{13}.3^9.17\)

\(=\left(2^6\right)^2.\left(3^4\right)^3.2.17\div2^{13}.3^9.17\)

\(=2^{12}.3^{12}.2.17\div2^{13}.3^9.17\)

\(=\left(2^{12}.2\div2^{13}\right).\left(3^{12}.3^9\right).\left(17.17\right)\)

\(=1.3^{21}.17^2\)

\(=3^{21}.17^2\)

21 tháng 8 2020

Bạn ơi nếu dấu chia kia là phân số thì làm theo cách dưới đây , còn không phải thì làm theo cách kia

\(\frac{64^2.81^3.34}{2^{13}.3^9.17}=\frac{\left(2^6\right)^2.\left(3^4\right)^3.2.17}{2^{13}.3^9.17}=\frac{2^{12}.3^{12}.2.17}{2^{13}.3^9.17}=\frac{2^{13}.3^{12}.17}{2^{13}.3^9.17}=3^3=27\)

20 tháng 8 2020

\(\frac{\left(2^6\right)^2.\left(3^4\right)^3.34}{2^{13}.3^9.17}=\frac{2^{12}.3^{12}.2}{2^{13}.3^9}=3^3=27\)

21 tháng 8 2020

thanks

12 tháng 5 2022

\(=\dfrac{\left(2^3\right)^3.\left(3^2\right)^4-2^8.\left(3^4\right)^2}{\left(2^4\right)^2.\left(3^4\right)^2+\left(2^2\right)^4.\left(3^3\right)^3}=\dfrac{2^9.3^8-2^8.3^8}{2^8.3^8+2^8.3^9}=\)

\(=\dfrac{2^8.3^8.\left(2-1\right)}{2^8.3^8.\left(1+3\right)}=\dfrac{1}{4}\)

4 tháng 5 2023

a/\(C=\dfrac{2}{1.7}+\dfrac{2}{7.13}+\dfrac{2}{13.19}+...+\dfrac{2}{1013.1019}\)
\(=\dfrac{1}{3}\left(\dfrac{6}{1.7}+\dfrac{6}{7.13}+\dfrac{6}{13.19}+...+\dfrac{6}{1013.1019}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+...+\dfrac{1}{1013}-\dfrac{1}{1019}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{1019}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{1018}{1019}\)
\(=\dfrac{1018}{3057}\)
b/\(D=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{2011.2019}\)
\(=\dfrac{7}{8}\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{2011.2019}\right)\)
\(=\dfrac{7}{8}\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{2011}-\dfrac{1}{2019}\right)\)
\(=\dfrac{7}{8}\left(1-\dfrac{1}{2019}\right)\)
\(=\dfrac{7}{8}\cdot\dfrac{2018}{2019}\)
\(=\dfrac{7063}{8076}\)

4 tháng 5 2023

Thankyou, Love youyeu

20 tháng 7 2019

1. Tìm x, biết :

a. ( x - \(\frac{3}{4}\)\(^2\)= 0

=> x - \(\frac{3}{4}\)= 0

=> x = 0 + \(\frac{3}{4}\)

=> x = \(\frac{3}{4}\)

b. ( x + \(\frac{1}{2}\)\(^2\)\(\frac{9}{64}\)

=> ( x + \(\frac{1}{2}\)\(^2\)= ( \(\frac{3}{8}\)\(^2\)

=> x + \(\frac{1}{2}\)\(\frac{3}{8}\)

=> x = \(\frac{3}{8}\)\(\frac{1}{2}\)

=> x = \(\frac{-1}{8}\)

c.  \(\frac{\left(-2\right)^x}{16}=-8\)

=> \(\frac{\left(-2\right)^x}{16}=\frac{-8}{1}=\frac{-128}{16}\)

=> ( -2)\(^x\)= -128

=> ( -2 ) \(^x\)= ( -2) \(^7\)

=> x = 7

29 tháng 6 2016

\(\frac{4^3\cdot9^3}{8^2\cdot81^2}=\frac{2^6\cdot3^6}{2^6\cdot3^8}=\frac{1}{3^2}=\frac{1}{9}\)

29 tháng 6 2016

\(\frac{4^3.9^3}{8^2.81^2}=\frac{\left(2^2\right)^3.\left(3^2\right)^3}{\left(2^3\right)^2.\left(3^4\right)^2}=\frac{2^6.3^6}{2^6.3^8}=\frac{1}{9}\)

\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}\)

\(=\dfrac{1}{2}\cdot\dfrac{3}{4}=\dfrac{3}{8}\)