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\(\frac{3^{10}\times11+3^5\times5}{3^9\times2^4}\)

\(\Rightarrow\)\(\frac{3^5\left(3^5\times11+5\right)}{3^5\times3^4\times2^4}\)

\(\Rightarrow\frac{3^5\times11+5}{\left(3\times2\right)^4}\)

\(\Rightarrow\frac{3^5\times11+5}{6^4}\)

\(\Rightarrow\frac{3^5\times11}{6^4}+\frac{5}{6^4}\)

\(\Rightarrow\frac{3^5\times11}{2^4\times3^4}+\frac{5}{6^4}\)

\(\Rightarrow\frac{3\times11}{16}+\frac{5}{1295}\)

\(\Rightarrow\frac{33}{16}+\frac{5}{1295}\)

\(\Rightarrow\frac{8563}{4144}\)

16 tháng 9 2017

\(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)

\(\Rightarrow\frac{2^{12}.\left(13+65\right)}{2^{10}.104}+\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\)

\(\Rightarrow\frac{2^{12}.78}{2^{10}.104}+\frac{3^{10}.2^4}{3^9.2^4}\)

\(=\frac{2^2.3}{4}+3\)

\(=3+3=6\)

18 tháng 7 2016

\(A=\frac{72^3\times54^2}{108^4}=\frac{\left(2^3\times3^2\right)^3\times\left(2\times3^3\right)^2}{\left(2^2\times3^3\right)^4}=\frac{2^9\times3^6\times2^2\times3^6}{2^8\times3^{12}}=2^3=8\)

\(B=\frac{3^{10}\times11+3^{10}\times5}{3^9\times2^4}=\frac{3^{10}\times\left(11+5\right)}{3^9\times16}=\frac{3\times16}{16}=3\)

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18 tháng 7 2016

giúp với mik cần gấp

18 tháng 6 2018

B=3^10.11+3^10.5/3^9.2^4

  = 3^10( 11+5)/3^9.16

  = 3^10.16/3^9.16

  = 3^10/3^9

  = 3

Vậy B = 3 (1)

C = 2^10.13+2^10.65/2^8.104

   = 2^10(13+65)/2^8.2^2.26

   = 2^10.78/2^10.26

   = 78/26

   = 3

Vậy C = 3 (2)

Từ (1) v (2) suy ra B=C

6 tháng 8 2016

\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)

\(=\frac{1}{1.3}-\frac{1}{11.13}\)

\(=\frac{1}{3}-\frac{1}{143}\)

\(=\frac{140}{429}\)

12 tháng 3 2019

Bài 5:

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}< 1\)

Vậy A<1.

Bài 4: Bn ghi nhầm đề rồi.

Đề đúng: \(A=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{2011.2013}\)

\(\frac{1}{2}A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2011.2013}\)

\(\frac{1}{2}A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\)

\(\frac{1}{2}A=1-\frac{1}{2013}\)

\(A=2.\frac{2012}{2013}=\frac{4024}{2013}\)

13 tháng 3 2019

-__-

23 tháng 9 2020

\(A=\frac{6^{10}-3^9.2^8.5}{27^3.4^5+16^3.9^4}\)

\(=\frac{3^{10}.2^{10}-3^9.2^8.5}{\left(3^3\right)^3.\left(2^2\right)^5+\left(2^4\right)^3.\left(3^2\right)^4}\)

\(=\frac{3^{10}.2^{10}-3^9.2^8.5}{3^9.2^{10}+2^{12}.3^8}\)

\(=\frac{3^9.2^8.\left(3.2^2-1.1.5\right)}{3^8.2^{10}.\left(3.1+2^2\right)}\)

\(=\frac{3^9.2^8.7}{3^8.2^{10}.7}\)

\(=\frac{3}{2^2}=\frac{3}{4}\)

Bài làm :

\(A=\frac{6^{10}-3^9.2^8.5}{27^3.4^5+16^3.9^4}\)

\(=\frac{\left(2.3\right)^{10}-3^9.2^8.5}{\left(3^3\right)^3.\left(2^2\right)^5+\left(2^4\right)^3.\left(3^2\right)^4}\)

\(=\frac{2^{10}.3^{10}-3^9.2^8.5}{3^9.2^{10}+2^{12}.3^8}\)

\(=\frac{2^8.3^9.\left(2^2.3-5\right)}{3^8.2^{10}.\left(3+2^2\right)}\)

\(=\frac{3.7}{2^2.7}\)

\(=\frac{3}{4}\)

Học tốt

9 tháng 10 2020

\(A=\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\frac{4}{9\cdot11}\)

\(A=\frac{2\cdot2}{3\cdot5}+\frac{2\cdot2}{5\cdot7}+\frac{2\cdot2}{7\cdot9}+\frac{2\cdot2}{9\cdot11}\)

\(A=2\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right)\)

\(A=2\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(A=2\cdot\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(A=2\cdot\frac{8}{33}\)

\(A=\frac{16}{33}\)

9 tháng 10 2020

Ta có: 

\(A=\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}\)

\(A=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(A=2\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(A=2\cdot\frac{8}{33}\)

\(A=\frac{16}{33}\)