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( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +1/5.6 ) x 10 - x = 0
= ( 1- 1/2 +1/2 -1/3 +1/3 - 1/4 + 1/4 - 1/5 +1/5 -1/6 ) x 10 - x = 0
= ( 1 - 1/6 ) x 10 - x = 0
= 5/6 x 10 - x =0
= 25/3 - x =0
x = 25/3 - 0
x = 25/3
\(\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\right)\times10-x=0\)
\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\times10-x=0\)
\(\left(\frac{1}{1}-\frac{1}{6}\right)\times10-x=0\)
\(\frac{5}{6}\times10-x=0\)
\(\frac{25}{3}-x=0\)
x =\(\frac{25}{3}-0=\frac{25}{3}\)
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a,Đặt \(A=\frac{1}{1\times4}+\frac{1}{4\times7}+...+\frac{1}{97\times100}\)
\(\Rightarrow3A=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{97\times100}\)
\(\Rightarrow3A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow3A=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{300}\)
b, \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}=\frac{1\times2\times...\times99}{2\times3\times...\times1000}=\frac{1}{100}\)
c, \(\frac{3}{4}\times\frac{8}{9}\times...\times\frac{99}{100}=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times...\times\frac{9.11}{10.10}=\frac{1.2.....9}{2.3.....10}\times\frac{3.4.....11}{2.3.....10}=\frac{1}{10}\times\frac{11}{2}=\frac{11}{20}\) (dấu . là dấu nhân)
\(\frac{1}{8}=12,5\%\) ; \(\frac{1}{16}=6,25\%\) ; \(\frac{1}{2}=50\%\) ; \(\frac{1}{4}=25\%\)
Thay vào trên mà tính.
= \(1+\left(\frac{3\left(1x2+2x4x2\right)}{3\left(5+5x3x25\right)}+1\right)-\left(1+\frac{18}{54}\right)-1\) = \(\frac{18}{380}-\frac{18}{54}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\right)\cdot x=2009\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\right)\cdot x=2009\)
\(\left(1-\frac{1}{2010}\right)\cdot x=2009\)
\(\frac{2009}{2010}\cdot x=2009\)
\(x=2009:\frac{2009}{2010}\)
\(x=2010\)
1/1x2 + 1/2x3 +1/3x4 + ......+1/98x99+1/99x100
=1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +......+ 1/98 - 1/99 + 1/99 + 1/100
=(1-1/100)+(1/2 - 1/2 ) + ( 1/3 - 1/3 ) + ...... + (1/98 - 1/98 ) + ( 1/99 - 1/99 )
= 100/100 - 1/100 + 0 + 0 +.....+ 0 + 0
=99/100
vậy GTBT = 99/100