Tìm n thuôc z, sao cho): a) n+1 trên n+2 thuôc Z (2 cách) b) 2n+3 trên 2n+1 thuộc Z 22/10/2021 Bởi Harper Tìm n thuôc z, sao cho): a) n+1 trên n+2 thuôc Z (2 cách) b) 2n+3 trên 2n+1 thuộc Z
a/ $\dfrac{n+1}{n+2}=\dfrac{n+2-1}{n+2}=1-\dfrac{1}{n+2}\in \mathbb Z$ $→1\vdots n+2$ $→n+2\in Ư(1)=\{±1\}$ $→n\in \{-1;-3\}$ b/ $\dfrac{2n+3}{2n+1}=\dfrac{2n+1+2}{2n+1}=1+\dfrac{2}{2n+1}∈\mathbb Z$ $→2\vdots 2n+1$ $→2n+1\in Ư(2)=\{±1;±2\}$ $→n\in \{0;-1;\dfrac{1}{2};-\dfrac{3}{2}\}$ mà $n\in \mathbb Z$ $→n\in \{0;-1\}$ Bình luận
Đáp án: ._. Giải thích các bước giải: `a)` Cách `1:` `(n+1)/(n+2) in ZZ` `<=> n+1 vdots n+2` `=> n+2-1 vdots n+2` `=> 1 vdots n+2` `=> n+2 in Ư(1)={-1;1}` `=> n in {-3;-1}` Cách `2:` `(n+1)/(n+2)=(n+2-1)/(n+2)=1-1/(n+2) in ZZ` `=> 1 vdots n+2` `=> n+2 in Ư(1)={-1;1}` `=> n in {-3;-1}` `b)` `(2n+3)/(2n+1) =(2n+1+2)/(2n+1)=1+2/(2n+1) in ZZ` `=> 2 vdots 2n+1` `=> 2n+1 in Ư(2)={-2;-1;1;2}` `=> 2n in {-3;-2;0;1}` `=> n in {-3/2;-1;0;1/2}` Mà `n in ZZ` `=> n in {-1;0}` Bình luận
a/ $\dfrac{n+1}{n+2}=\dfrac{n+2-1}{n+2}=1-\dfrac{1}{n+2}\in \mathbb Z$
$→1\vdots n+2$
$→n+2\in Ư(1)=\{±1\}$
$→n\in \{-1;-3\}$
b/ $\dfrac{2n+3}{2n+1}=\dfrac{2n+1+2}{2n+1}=1+\dfrac{2}{2n+1}∈\mathbb Z$
$→2\vdots 2n+1$
$→2n+1\in Ư(2)=\{±1;±2\}$
$→n\in \{0;-1;\dfrac{1}{2};-\dfrac{3}{2}\}$ mà $n\in \mathbb Z$
$→n\in \{0;-1\}$
Đáp án:
._.
Giải thích các bước giải:
`a)`
Cách `1:`
`(n+1)/(n+2) in ZZ`
`<=> n+1 vdots n+2`
`=> n+2-1 vdots n+2`
`=> 1 vdots n+2`
`=> n+2 in Ư(1)={-1;1}`
`=> n in {-3;-1}`
Cách `2:`
`(n+1)/(n+2)=(n+2-1)/(n+2)=1-1/(n+2) in ZZ`
`=> 1 vdots n+2`
`=> n+2 in Ư(1)={-1;1}`
`=> n in {-3;-1}`
`b)`
`(2n+3)/(2n+1) =(2n+1+2)/(2n+1)=1+2/(2n+1) in ZZ`
`=> 2 vdots 2n+1`
`=> 2n+1 in Ư(2)={-2;-1;1;2}`
`=> 2n in {-3;-2;0;1}`
`=> n in {-3/2;-1;0;1/2}`
Mà `n in ZZ`
`=> n in {-1;0}`