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10 tháng 7 2015

ta có :\(\left(1-\frac{1}{5}\right)x\left(1-\frac{1}{6}\right)x\left(1-\frac{1}{7}\right)x\left(1-\frac{1}{8}\right)\)

\(=\frac{4}{5}x\frac{5}{6}x\frac{6}{7}x\frac{7}{8}=\frac{4}{8}=\frac{1}{2}\)

10 tháng 7 2015

\(\left(1-\frac{1}{5}\right).\left(1-\frac{1}{6}\right).\left(1-\frac{1}{7}\right).\left(1-\frac{1}{8}\right)\)

\(=\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}\)

\(=\frac{4}{8}\)

\(=\frac{1}{2}\)

2 tháng 5 2017

1/10x(2/9x9/2)x(3/8x8/3)x(5/6x6/5)=1/10x1x1x1=1/10

29 tháng 1 2018

a)(11*9-100+1):(1*2*3*4*5*6*7*8*9*10)

=0:(1*2*3*4*5*6*7*8*9*10)

=0

27 tháng 2 2020

a)(11*9-100+1):(1*2*3*4*5*6*7*8*9*10)
=0:(1*2*3*4*5*6*7*8*9*10)
=0

hacker 2k6

17 tháng 2 2016

minh ra 2 nhung chac khong dung

22 tháng 4 2018

\(=\frac{1}{5}\times\frac{1}{3}\times\frac{3}{7}\times\frac{1}{2}\)

\(=\left(\frac{1}{5}\times\frac{1}{2}\right)\times\left(\frac{1}{3}\times\frac{3}{7}\right)\)

\(=\frac{1}{10}\times\frac{1}{7}\)

\(=\frac{1}{70}\)

13 tháng 8 2017

\(\frac{2}{5}:\frac{3}{7}x\frac{3}{7}:\frac{2}{5}+1999=\)\(\frac{2}{5}x\frac{7}{3}x\frac{3}{7}x\frac{5}{2}+1999=\)\(\left(\frac{2x7x3x5}{5x3x7x2}\right)+1999=\)\(1+1999=2000\)

\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)\)\(=\left(\frac{2}{2}-\frac{1}{2}\right)\)\(x\left(\frac{3}{3}-\frac{1}{3}\right)\)\(x\left(\frac{4}{4}-\frac{1}{4}\right)\)\(x\left(\frac{5}{5}-\frac{1}{5}\right)\)\(=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}=\)\(\frac{1x2x3x4}{2x3x4x5}=\frac{1x\left(2x3x4\right)}{\left(2x3x4\right)x5}=\frac{1}{5}\)\(.\)

13 tháng 8 2017

2/5:3/7 nhân 3/7:2/5+1999

=2/5 nhân 7/3 nhân 3/7nhân 5/2+1999

=14/15 nhân 15/14+1999

=1+1999

=2000

(1-1/2) nhân (1-1/3) nhân (1-1/4) nhân (1-1/5)

=1/2 nhân 2/3 nhân 3/4 nhân 4/5

=24/120

=1/5

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)

\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{20}{41}\)

\(\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{\left(x+2\right)-x}{x\left(x+2\right)}\right)=\frac{20}{41}\)

\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)

\(1-\frac{1}{x+2}=\frac{40}{41}\)

\(\frac{1}{x+2}=1-\frac{40}{41}\)

\(\frac{1}{x+2}=\frac{1}{41}\)

\(x+2=41\)

\(x=41-2\)

\(x=39\)

24 tháng 8 2019

Tìm x 

a) \(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{x\times\left(x+2\right)}=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+1\right)}\right)=\frac{20}{41}\)

\(\Rightarrow\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+2\right)}=\frac{20}{41}:\frac{1}{2}\)

\(\Rightarrow\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+2\right)}=\frac{40}{41}\)

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{\left(x+2\right)}=\frac{40}{41}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{41}\)

\(\Rightarrow x+2=41\)

\(\Rightarrow x=41-2\)

\(\Rightarrow x=39\)

Vậy x = 39