Toán chứng tỏ rằng 1 phần 101 + 1 phần 102 + 1 phần 103 +…+1 phần 300 > 2 phần 3 08/10/2021 By Melanie chứng tỏ rằng 1 phần 101 + 1 phần 102 + 1 phần 103 +…+1 phần 300 > 2 phần 3
Ta đặt `A=1/101+1/102+1/103+…+1/300` `⇒A=(1/101+1/102+…+1/200)+(1/201+1/202+…+1/300)` `⇒A>(1/200+1/200+…+1/200)+(1/201+1/202+…+1/300)` `⇒A>(\underbrace{1/200+1/200+…+1/200}_{\text{100 chữ số}})+(\underbrace{1/300+1/300+…+1/300}_{\text{100 chữ số}})` `⇒A>1/200 . 100+1/300 . 100` `⇒A>100/200+100/300` `⇒A>1/2+1/3` `⇒A>5/6>4/6=2/3` `⇒A>2/3` Trả lời
$\dfrac{1}{101} + \dfrac{1}{102} + \dfrac{1}{103} + …. + \dfrac{1}{300}$ $= (\dfrac{1}{101} + \dfrac{1}{102} + \dfrac{1}{103} + …. + \dfrac{1}{200}) + (\dfrac{1}{201} + \dfrac{1}{202} + … + \dfrac{1}{300})$ $\dfrac{1}{101} + \dfrac{1}{102} + \dfrac{1}{103} + …. + \dfrac{1}{200} > \dfrac{1}{200} + \dfrac{1}{200} + \dfrac{1}{200} + …. + \dfrac{1}{200}$ $\dfrac{1}{201} + \dfrac{1}{202} + \dfrac{1}{203} + …. + \dfrac{1}{300} > \dfrac{1}{300} + \dfrac{1}{300} + \dfrac{1}{300} + …. + \dfrac{1}{300}$ $⇒ (\dfrac{1}{101} + \dfrac{1}{102} + \dfrac{1}{103} + …. + \dfrac{1}{200}) + (\dfrac{1}{201} + \dfrac{1}{202} + … + \dfrac{1}{300}) > (\dfrac{1}{200} + \dfrac{1}{200} + \dfrac{1}{200} + …. + \dfrac{1}{200}) + (\dfrac{1}{300} + \dfrac{1}{300} + \dfrac{1}{300} + …. + \dfrac{1}{300}) = \dfrac{1}{200} . 100 + \dfrac{1}{300} . 100 = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6} > \dfrac{4}{6} = \dfrac{2}{3}$ $⇒ A > \dfrac{2}{3}$ ($đpcm$) Trả lời
Ta đặt `A=1/101+1/102+1/103+…+1/300`
`⇒A=(1/101+1/102+…+1/200)+(1/201+1/202+…+1/300)`
`⇒A>(1/200+1/200+…+1/200)+(1/201+1/202+…+1/300)`
`⇒A>(\underbrace{1/200+1/200+…+1/200}_{\text{100 chữ số}})+(\underbrace{1/300+1/300+…+1/300}_{\text{100 chữ số}})`
`⇒A>1/200 . 100+1/300 . 100`
`⇒A>100/200+100/300`
`⇒A>1/2+1/3`
`⇒A>5/6>4/6=2/3`
`⇒A>2/3`
$\dfrac{1}{101} + \dfrac{1}{102} + \dfrac{1}{103} + …. + \dfrac{1}{300}$
$= (\dfrac{1}{101} + \dfrac{1}{102} + \dfrac{1}{103} + …. + \dfrac{1}{200}) + (\dfrac{1}{201} + \dfrac{1}{202} + … + \dfrac{1}{300})$
$\dfrac{1}{101} + \dfrac{1}{102} + \dfrac{1}{103} + …. + \dfrac{1}{200} > \dfrac{1}{200} + \dfrac{1}{200} + \dfrac{1}{200} + …. + \dfrac{1}{200}$
$\dfrac{1}{201} + \dfrac{1}{202} + \dfrac{1}{203} + …. + \dfrac{1}{300} > \dfrac{1}{300} + \dfrac{1}{300} + \dfrac{1}{300} + …. + \dfrac{1}{300}$
$⇒ (\dfrac{1}{101} + \dfrac{1}{102} + \dfrac{1}{103} + …. + \dfrac{1}{200}) + (\dfrac{1}{201} + \dfrac{1}{202} + … + \dfrac{1}{300}) > (\dfrac{1}{200} + \dfrac{1}{200} + \dfrac{1}{200} + …. + \dfrac{1}{200}) + (\dfrac{1}{300} + \dfrac{1}{300} + \dfrac{1}{300} + …. + \dfrac{1}{300}) = \dfrac{1}{200} . 100 + \dfrac{1}{300} . 100 = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6} > \dfrac{4}{6} = \dfrac{2}{3}$
$⇒ A > \dfrac{2}{3}$ ($đpcm$)