Toán tìm x biết 1/5.8+1/8.11+1/11.14+…+1/x+(x+3)=134/2015 18/11/2021 By Valerie tìm x biết 1/5.8+1/8.11+1/11.14+…+1/x+(x+3)=134/2015
Ta có công thức : `a/{n(n+a)}=1/n-1/{n+1}` Trở lại với bài toán : `1/5.8+1/8.11+1/11.14+…+1/{x+(x+3)}=134/2015` `⇒3/5.8+3/8.11+3/11.14+…+3/{x+(x+3)}=402/2015` Áp dụng công thức đầu bài : `⇒1/5-1/8+1/8-1/11+1/11-1/14+…+1/x-1/{x+3}=402/2015` `⇒1/5-1/{x+3}=402/2015` `⇒{x+3-5}/{5(x+3)}=402/2015` `2015(x-2)=402(5x+15)` `2015x-4030=2010x+6030` `2015x-2010x=6030+4030` `5x=10060` `x=2012` Trả lời
$\dfrac{1}{5.8}+\dfrac{1}{8.11}+….+\dfrac{1}{x.(x+3)}=\dfrac{134}{2015}$ $\to \dfrac{1}{3}.\bigg(\dfrac{3}{5.8}+\dfrac{3}{8.11}+….+\dfrac{1}{x.(x+3)} \bigg) = \dfrac{134}{2015}$ $\to \dfrac{1}{5} – \dfrac{1}{8}+\dfrac{1}{8} – …..+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{402}{2015}$ $\to \dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{402}{2015}$ $\to \dfrac{x-2}{x+3} = \dfrac{402}{403}$ $\to 403x-806=402x+1206$ $\to x = 2012$ Vậy $x=2012$ Trả lời
Ta có công thức : `a/{n(n+a)}=1/n-1/{n+1}`
Trở lại với bài toán :
`1/5.8+1/8.11+1/11.14+…+1/{x+(x+3)}=134/2015`
`⇒3/5.8+3/8.11+3/11.14+…+3/{x+(x+3)}=402/2015`
Áp dụng công thức đầu bài :
`⇒1/5-1/8+1/8-1/11+1/11-1/14+…+1/x-1/{x+3}=402/2015`
`⇒1/5-1/{x+3}=402/2015`
`⇒{x+3-5}/{5(x+3)}=402/2015`
`2015(x-2)=402(5x+15)`
`2015x-4030=2010x+6030`
`2015x-2010x=6030+4030`
`5x=10060`
`x=2012`
$\dfrac{1}{5.8}+\dfrac{1}{8.11}+….+\dfrac{1}{x.(x+3)}=\dfrac{134}{2015}$
$\to \dfrac{1}{3}.\bigg(\dfrac{3}{5.8}+\dfrac{3}{8.11}+….+\dfrac{1}{x.(x+3)} \bigg) = \dfrac{134}{2015}$
$\to \dfrac{1}{5} – \dfrac{1}{8}+\dfrac{1}{8} – …..+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{402}{2015}$
$\to \dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{402}{2015}$
$\to \dfrac{x-2}{x+3} = \dfrac{402}{403}$
$\to 403x-806=402x+1206$
$\to x = 2012$
Vậy $x=2012$