Tìm x biết: `\frac{1}{3}“\frac{1}{6}+“\frac{1}{10}+…+“\frac{1}{x(x+1):2}=“\frac{2013}{2015}` 07/10/2021 Bởi Ximena Tìm x biết: `\frac{1}{3}“\frac{1}{6}+“\frac{1}{10}+…+“\frac{1}{x(x+1):2}=“\frac{2013}{2015}`
`\frac{1}{3}+“\frac{1}{6}+“\frac{1}{10}+…+“\frac{1}{x(x+1):2}=“\frac{2013}{2015}` `\frac{2}{6}+“\frac{2}{12}+“\frac{2}{20}+…+“\frac{2}{x(x+1)}=“\frac{2013}{2015}` `2“(“\frac{1}{2.3}+“\frac{1}{3.4}+“\frac{1}{4.5}+…+“\frac{1}{x(x+1)})=“\frac{2013}{2015}` `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{2013}{2015}` `2(“\frac{1}{2}-“\frac{1}{x+1})=“\frac{2013}{2015}` `\frac{1}{2}-“\frac{1}{x+1}=“\frac{2013}{2015}:2` `\frac{1}{2}-“\frac{1}{x+1}=“\frac{2013}{2015.2}` `\frac{1}{2}-“\frac{2013}{4030}=“\frac{1}{x+1}` `\frac{2015-2013}{4030}=“\frac{1}{x+1}` `\frac{1}{2015}=“\frac{1}{x+1}` `2015=x+1` `x=2015-1` `x=2014` Bình luận
Đáp án:
$x=2014$
Giải thích các bước giải:
`\frac{1}{3}+“\frac{1}{6}+“\frac{1}{10}+…+“\frac{1}{x(x+1):2}=“\frac{2013}{2015}`
`\frac{2}{6}+“\frac{2}{12}+“\frac{2}{20}+…+“\frac{2}{x(x+1)}=“\frac{2013}{2015}`
`2“(“\frac{1}{2.3}+“\frac{1}{3.4}+“\frac{1}{4.5}+…+“\frac{1}{x(x+1)})=“\frac{2013}{2015}`
`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{2013}{2015}`
`2(“\frac{1}{2}-“\frac{1}{x+1})=“\frac{2013}{2015}`
`\frac{1}{2}-“\frac{1}{x+1}=“\frac{2013}{2015}:2`
`\frac{1}{2}-“\frac{1}{x+1}=“\frac{2013}{2015.2}`
`\frac{1}{2}-“\frac{2013}{4030}=“\frac{1}{x+1}`
`\frac{2015-2013}{4030}=“\frac{1}{x+1}`
`\frac{1}{2015}=“\frac{1}{x+1}`
`2015=x+1`
`x=2015-1`
`x=2014`