Toán Tìm n để $\frac{n+3}{2n-5}$ là số nguyên. 17/09/2021 By Peyton Tìm n để $\frac{n+3}{2n-5}$ là số nguyên.
Để $\dfrac{n+3}{2n-5}$ $∈$ $Z$ thì : $n+3 \vdots 2n-5$ $⇔ 2.(n+3) – (2n-5) \vdots 2n-5$ $⇔ 2n + 6 – 2x + 5 \vdots 2n-5$ $⇔ 11 \vdots 2n-5$ $⇒ 2n-5$ $∈$ `Ư(11)={±1;±11}` $⇔2n$ $∈$ `{-6;4;6;16}` $⇔n$ $∈$ `{-3;2;3;8}` Vậy $n$ $∈$ `{-3;2;3;8}` thì $\dfrac{n+3}{2n-5}$ $∈$ $Z$. Trả lời
`(n+3)/(2n-5)` nhận giá trị nguyên khi $n+3$ ⋮ $2n-5$ $⇒2n+6$ ⋮ $2n-5$ $⇒(2n-5)+11$ ⋮ $2n-5$ $⇒11$ ⋮ $2n-5$ $⇒2n-5∈Ư(11)=\{±1;±11\}$ Ta có bảng sau: 2n-5 -1 1 -11 11 n 2 3 -3 8 Vậy $n∈\{2;3;-3;8\}$. Trả lời
Để $\dfrac{n+3}{2n-5}$ $∈$ $Z$ thì : $n+3 \vdots 2n-5$
$⇔ 2.(n+3) – (2n-5) \vdots 2n-5$
$⇔ 2n + 6 – 2x + 5 \vdots 2n-5$
$⇔ 11 \vdots 2n-5$
$⇒ 2n-5$ $∈$ `Ư(11)={±1;±11}`
$⇔2n$ $∈$ `{-6;4;6;16}`
$⇔n$ $∈$ `{-3;2;3;8}`
Vậy $n$ $∈$ `{-3;2;3;8}` thì $\dfrac{n+3}{2n-5}$ $∈$ $Z$.
`(n+3)/(2n-5)` nhận giá trị nguyên khi $n+3$ ⋮ $2n-5$
$⇒2n+6$ ⋮ $2n-5$
$⇒(2n-5)+11$ ⋮ $2n-5$
$⇒11$ ⋮ $2n-5$
$⇒2n-5∈Ư(11)=\{±1;±11\}$
Ta có bảng sau:
2n-5 -1 1 -11 11
n 2 3 -3 8
Vậy $n∈\{2;3;-3;8\}$.