a)Cho A=1/1×2+1/3×4+1/5×6+…..+1/99×100 B=2013/51+2013/52+2013/53+…..+2013/100 29/08/2021 Bởi Josephine a)Cho A=1/1×2+1/3×4+1/5×6+…..+1/99×100 B=2013/51+2013/52+2013/53+…..+2013/100
Đáp án: ( đpcm ) Giải thích các bước giải: `A= 1/1.2 + 1/3.4 + 1/5.6 +…+1/99.100` `= 1 -1/2 + 1/3 – 1/4 +1/5- 1/6+…+1/99-1/100` `= (1+1/3 + 1/5+…+1/99) – (1/2 + 1/4 +…+1/100)` `= (1+1/2 +1/3 +1/4+…+1/99 +1/100) – 2(1/2 + 1/4 +…+1/100)` `= 1/51 + 1/52 +…+1/100` $\\$ `B=2013/51 + 2013/52 + … + 2013/100` `=2013 . (1/51 + 1/52 +…+1/100)` $\\$ $\to \rm \dfrac{B}{A}= \dfrac{\dfrac{2013}{51} + \dfrac{2013}{52} +…+\dfrac{2013}{100}}{ \dfrac{1}{51}+\dfrac{1}{52}+…+\dfrac{1}{100}} = \dfrac{2013(\dfrac{1}{51}+\dfrac{1}{52}+…+\dfrac{1}{100})}{\dfrac{1}{51}+\dfrac{1}{52}+…+\dfrac{1}{100}}$ `= 2013` Vậy `B/A` có giá trị là `1` số nguyên Bình luận
Ta có: `A= 1/1.2 + 1/3.4 + 1/5.6 +…+1/99.100` `A= 1 -1/2 + 1/3 – 1/4 +1/5- 1/6+…+1/99-1/100` `A= (1+1/3 + 1/5+…+1/99) – (1/2 + 1/4 +…+1/100)` `A= (1+1/2 +1/3 +1/4+…+1/99 +1/100) – 2(1/2 + 1/4 +…+1/100)` `A= 1+1/2 +1/3 + 1/4+…+1/99 + 1/100 – 1 – 1/2 -…-1/50` `A= 1/51 + 1/52 +…+1/100` `=> B/A = (2013/51 + 2013/52 + 2013/53+…+2013/100) / ( 1/51 +1/52+…+1/100)` `=> B/A = (2013(1/51 + 1/52+ 1/53+…+1/100))/(1/51 +1/52 +…+1/100)` `=> B/A = 2013 in Z` Vậy `B/A` có giá trị là 1 số nguyên Bình luận
Đáp án:
( đpcm )
Giải thích các bước giải:
`A= 1/1.2 + 1/3.4 + 1/5.6 +…+1/99.100`
`= 1 -1/2 + 1/3 – 1/4 +1/5- 1/6+…+1/99-1/100`
`= (1+1/3 + 1/5+…+1/99) – (1/2 + 1/4 +…+1/100)`
`= (1+1/2 +1/3 +1/4+…+1/99 +1/100) – 2(1/2 + 1/4 +…+1/100)`
`= 1/51 + 1/52 +…+1/100`
$\\$
`B=2013/51 + 2013/52 + … + 2013/100`
`=2013 . (1/51 + 1/52 +…+1/100)`
$\\$
$\to \rm \dfrac{B}{A}= \dfrac{\dfrac{2013}{51} + \dfrac{2013}{52} +…+\dfrac{2013}{100}}{ \dfrac{1}{51}+\dfrac{1}{52}+…+\dfrac{1}{100}} = \dfrac{2013(\dfrac{1}{51}+\dfrac{1}{52}+…+\dfrac{1}{100})}{\dfrac{1}{51}+\dfrac{1}{52}+…+\dfrac{1}{100}}$
`= 2013`
Vậy `B/A` có giá trị là `1` số nguyên
Ta có: `A= 1/1.2 + 1/3.4 + 1/5.6 +…+1/99.100`
`A= 1 -1/2 + 1/3 – 1/4 +1/5- 1/6+…+1/99-1/100`
`A= (1+1/3 + 1/5+…+1/99) – (1/2 + 1/4 +…+1/100)`
`A= (1+1/2 +1/3 +1/4+…+1/99 +1/100) – 2(1/2 + 1/4 +…+1/100)`
`A= 1+1/2 +1/3 + 1/4+…+1/99 + 1/100 – 1 – 1/2 -…-1/50`
`A= 1/51 + 1/52 +…+1/100`
`=> B/A = (2013/51 + 2013/52 + 2013/53+…+2013/100) / ( 1/51 +1/52+…+1/100)`
`=> B/A = (2013(1/51 + 1/52+ 1/53+…+1/100))/(1/51 +1/52 +…+1/100)`
`=> B/A = 2013 in Z`
Vậy `B/A` có giá trị là 1 số nguyên