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Sửa đề \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2016}\div2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4032}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4032}\Leftrightarrow\frac{1}{x+1}=\frac{1}{4032}\)
\(\Leftrightarrow x+1=4032\Rightarrow x=4031\)
\(=\frac{3}{1}.\frac{4}{2}.\frac{5}{3}...\frac{2018}{2016}.\frac{2019}{2017}\\ =\frac{3.4.5...2018.2019}{1.2.3...2016.2017}\\ =\frac{2018.2019}{2}=1009.2019\)
=>
\(\frac{3}{2}x\frac{4}{3}x\frac{5}{4}x...x\frac{2016}{2015}=\frac{3x4x5x...x2016}{2x3x4x....x2015}=\frac{2016}{2}=1008\)
vay C=1008
\(1\frac{1}{2}x1\frac{1}{3}x1\frac{1}{4}x..........x1\frac{1}{2015}\)
\(=\frac{3}{2}x\frac{4}{3}x\frac{5}{4}x.........x\frac{2016}{2015}\)
\(=\frac{2016}{2}=1008\)
\(\frac{1}{6}.\frac{1}{3}+\frac{17}{6}.\frac{1}{3}+\frac{2015}{2016}-1\)
\(=\frac{1}{3}\left(\frac{1}{6}+\frac{17}{6}\right)+\frac{2015}{2016}-1\)
\(=\frac{1}{3}.3+\frac{2015}{2016}-1\)
\(=1-1+\frac{2015}{2016}=\frac{2015}{2016}\)
\(\frac{1}{6}\times\frac{1}{3}+\frac{17}{6}\times\frac{1}{3}+\frac{2015}{2016}-1\)
\(=\left(\frac{1}{6}+\frac{17}{6}\right)\times\frac{1}{3}+\frac{2015}{2016}-1\)
\(=3\times\frac{1}{3}+\frac{2015}{2016}-1\)
\(=1+\frac{2015}{2016}-1\)
\(=0+\frac{2015}{2016}=\frac{2015}{2016}\)
=\(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2013}{2014}x\frac{2014}{2015}\)
=\(\frac{1x2x3x...x2013x2014}{2x3x4x...x2014x2015}\)
=\(\frac{1}{2015}\)
( Dau x la dau nhan)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2015}\right)\)
\(=\left(\frac{2-1}{2}\right)\left(\frac{3-1}{3}\right)\left(\frac{4-1}{4}\right)....\left(\frac{2015-1}{2015}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2013}{2014}.\frac{2014}{2015}\)
\(=\frac{1}{2015}\)
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{2015}\right)\times\left(1-\frac{1}{2016}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{2014}{2015}\times\frac{2015}{2016}\)
\(=\frac{1}{2016}\)
Giải : Ta có (1-1/2)*(1-1/3)*(1-1/4)*....*(1-1/2015)*(1-1/2016)
= 1* -(1/2+1/3+1/4+....+1/2015+1/2016)
= 1* - (1/2+1/2016 +1/3+1/2015 +...+1/1007)
= 1* -(1/2033134)
= -1/2033134