2/2.3 + 2/3.4 + 2/4.5 + … + 2/x(x+1) = 2007/2009 15/09/2021 Bởi Ariana 2/2.3 + 2/3.4 + 2/4.5 + … + 2/x(x+1) = 2007/2009
`2/2.3+2/3.4+2/4.5+…+2/{x(x+1)}=2007/2009` `⇒2[1/2.3+1/3.4+1/4.5+…+1/{x(x+1)}]=2007/2009` `⇒2(1/2-1/3+1/3-1/4+1/4-1/5+…+1/x-1/{x+1})=2007/2009` `⇒2(1/2-1/{x+1})=2007/2009` `⇒1-2/{x+1}=2007/2009` `⇒2/{x+1}=2/2009` `⇒x+1=2009` `⇒x=2008` Bình luận
Ptrinh tương đương vs $2(\dfrac{1}{2.3} + \dfrac{1}{3.4} + \cdots + \dfrac{1}{x(x+1)}) = \dfrac{2007}{2009}$ $<-> \dfrac{1}{2} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} + \cdots + \dfrac{1}{x} – \dfrac{1}{x+1} = \dfrac{2007}{4018}$ $<-> \dfrac{1}{2} – \dfrac{1}{x+1} = \dfrac{2007}{4018}$ $<-> \dfrac{1}{x+1} = \dfrac{1}{2} – \dfrac{2007}{4018}$ $<-> \dfrac{1}{x+1} = \dfrac{1}{2009}$ $<-> x+1 = 2009$ $<-> x = 2008$. Vậy nghiệm là 2008. Bình luận
`2/2.3+2/3.4+2/4.5+…+2/{x(x+1)}=2007/2009`
`⇒2[1/2.3+1/3.4+1/4.5+…+1/{x(x+1)}]=2007/2009`
`⇒2(1/2-1/3+1/3-1/4+1/4-1/5+…+1/x-1/{x+1})=2007/2009`
`⇒2(1/2-1/{x+1})=2007/2009`
`⇒1-2/{x+1}=2007/2009`
`⇒2/{x+1}=2/2009`
`⇒x+1=2009`
`⇒x=2008`
Ptrinh tương đương vs
$2(\dfrac{1}{2.3} + \dfrac{1}{3.4} + \cdots + \dfrac{1}{x(x+1)}) = \dfrac{2007}{2009}$
$<-> \dfrac{1}{2} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} + \cdots + \dfrac{1}{x} – \dfrac{1}{x+1} = \dfrac{2007}{4018}$
$<-> \dfrac{1}{2} – \dfrac{1}{x+1} = \dfrac{2007}{4018}$
$<-> \dfrac{1}{x+1} = \dfrac{1}{2} – \dfrac{2007}{4018}$
$<-> \dfrac{1}{x+1} = \dfrac{1}{2009}$
$<-> x+1 = 2009$
$<-> x = 2008$.
Vậy nghiệm là 2008.