1/Tìm x: x/-2=-8/x 2/Tìm n ∈ N* để n/n+1+2/n+1 13/10/2021 Bởi Katherine 1/Tìm x: x/-2=-8/x 2/Tìm n ∈ N* để n/n+1+2/n+1
`1)x/(-2)=(-8)/x` `=>x.x=(-2).(-8)` `=>x^2=16` `=>x^2=4^2=(-4)^2` `=>x∈{±4}` Vậy `x∈{±4}` `b)n/(n+1)+2/(n+1)=(n+2)/(n+1)=1+ 1/(n+1)` Để `n/(n+1)+2/(n+1) ∈Z` `⇒1/(n+1)∈Z` `⇒n+1∈Ư(1)={±1}` `⇒n∈{0,-2}` Mà `n∈N**⇒n≥1` `⇒n∈∅` Vậy `n∈∅` Bình luận
`1// x/-2=(-8)/x` `=>x^2=16` `=>x=+-4` `2//n/(n+1)+2/(n+1)inZZ` `=> (n+2)/(n+1)inZZ` `=>1/(n+1)inZZ` `=>1\vdotsn+1` `=>n+1in Ư_((1))={+-1}` `=>n={-2;0}` Vì `ninNN^**` `=>`$n\in\varnothing$ Bình luận
`1)x/(-2)=(-8)/x`
`=>x.x=(-2).(-8)`
`=>x^2=16`
`=>x^2=4^2=(-4)^2`
`=>x∈{±4}`
Vậy `x∈{±4}`
`b)n/(n+1)+2/(n+1)=(n+2)/(n+1)=1+ 1/(n+1)`
Để `n/(n+1)+2/(n+1) ∈Z`
`⇒1/(n+1)∈Z`
`⇒n+1∈Ư(1)={±1}`
`⇒n∈{0,-2}`
Mà `n∈N**⇒n≥1`
`⇒n∈∅`
Vậy `n∈∅`
`1// x/-2=(-8)/x`
`=>x^2=16`
`=>x=+-4`
`2//n/(n+1)+2/(n+1)inZZ`
`=> (n+2)/(n+1)inZZ`
`=>1/(n+1)inZZ`
`=>1\vdotsn+1`
`=>n+1in Ư_((1))={+-1}`
`=>n={-2;0}`
Vì `ninNN^**`
`=>`$n\in\varnothing$