Tim x biet; 1/3 cong 1/6 cong 1/10 cong……………. cong 2/x[x cong 1] = 2011/2013 . giai ro cac buoc 22/08/2021 Bởi Vivian Tim x biet; 1/3 cong 1/6 cong 1/10 cong……………. cong 2/x[x cong 1] = 2011/2013 . giai ro cac buoc
`\frac{1}{3} + \frac{1}{6} + \frac{1}{10} + … + \frac{1}{x(x+1)} = \frac{2011}{2013}` `=> \frac{2}{6} + \frac{2}{12} + \frac{2}{20} + … + \frac{2}{x(x+1)} = \frac{2011}{2013}` `=> 2( \frac{1}{2.3} + \frac{1}{3.4} + \frac{4.5} + … + \frac{1}{x(x+1)} ) = \frac{2011}{2013}` `=> 2( \frac{1}{2} – \frac{1}{3} + \frac{1}{3} – \frac{1}{4} + \frac{1}{4} – \frac{1}{5} + … + \frac{1}{x} – \frac{1}{x + 1} ) = \frac{2011}{2013}` `=> 2( \frac{1}{2} – \frac{1}{x+1} ) = \frac{2011}{2013}` `=> \frac{1}{2} – \frac{1}{x+1} = \frac{2011}{2013} : 2` `=> \frac{1}{2} – \frac{1}{x+1} = \frac{2011}{4026}` `=> \frac{1}{x+1} = \frac{1}{2} – \frac{2011}{4026}` `=> \frac{1}{x+1} = \frac{2}{4026}` `=> \frac{1}{x+1} = \frac{1}{2013}` `=> x + 1 = 2013` `=> x = 2012` Mình nghĩ là đề sai chỗ tử của `\frac{2}{x(x+1)}` nhé Bình luận
`\frac{1}{3} + \frac{1}{6} + \frac{1}{10} + … + \frac{1}{x(x+1)} = \frac{2011}{2013}`
`=> \frac{2}{6} + \frac{2}{12} + \frac{2}{20} + … + \frac{2}{x(x+1)} = \frac{2011}{2013}`
`=> 2( \frac{1}{2.3} + \frac{1}{3.4} + \frac{4.5} + … + \frac{1}{x(x+1)} ) = \frac{2011}{2013}`
`=> 2( \frac{1}{2} – \frac{1}{3} + \frac{1}{3} – \frac{1}{4} + \frac{1}{4} – \frac{1}{5} + … + \frac{1}{x} – \frac{1}{x + 1} ) = \frac{2011}{2013}`
`=> 2( \frac{1}{2} – \frac{1}{x+1} ) = \frac{2011}{2013}`
`=> \frac{1}{2} – \frac{1}{x+1} = \frac{2011}{2013} : 2`
`=> \frac{1}{2} – \frac{1}{x+1} = \frac{2011}{4026}`
`=> \frac{1}{x+1} = \frac{1}{2} – \frac{2011}{4026}`
`=> \frac{1}{x+1} = \frac{2}{4026}`
`=> \frac{1}{x+1} = \frac{1}{2013}`
`=> x + 1 = 2013`
`=> x = 2012`
Mình nghĩ là đề sai chỗ tử của `\frac{2}{x(x+1)}` nhé