o_O khó quá ạ :) $A=\dfrac{8n-9}{2n+5}$ tìm n để A nguyên. 22/09/2021 Bởi Liliana o_O khó quá ạ 🙂 $A=\dfrac{8n-9}{2n+5}$ tìm n để A nguyên.
$A=\frac{8n-9}{2n+5}=$ $\frac{4(2n+5)-29}{2n+5}=4-$ $\frac{29}{2n+5}$ Để $A$ nguyên $⇒\frac{29}{2n+5}∈Z⇒2n+5∈Ư(29)=\{±1;±29\}$ Ta có bảng tương ứng: 2n+5 1 -1 29 -29 2n -4 -6 24 -34 n -2 -3 12 -17 Vậy $x∈\{-2;-3;12;-17\}$ Bình luận
$A=\frac{8n-9}{2n+5}$ nguyên khi $8n-9$ ⋮ $2n+5$ $→(8n+20)+29$ ⋮ $2n+5$ $→4(2n+5)+29$ ⋮ $2n+5$ $→29$ ⋮ $2n+5$ $→2n+5∈Ư(29)=\{±1;±29\}$ $→n∈\{-2;-3;-17;12\}$ Bình luận
$A=\frac{8n-9}{2n+5}=$ $\frac{4(2n+5)-29}{2n+5}=4-$ $\frac{29}{2n+5}$
Để $A$ nguyên $⇒\frac{29}{2n+5}∈Z⇒2n+5∈Ư(29)=\{±1;±29\}$
Ta có bảng tương ứng:
2n+5 1 -1 29 -29
2n -4 -6 24 -34
n -2 -3 12 -17
Vậy $x∈\{-2;-3;12;-17\}$
$A=\frac{8n-9}{2n+5}$ nguyên khi $8n-9$ ⋮ $2n+5$
$→(8n+20)+29$ ⋮ $2n+5$
$→4(2n+5)+29$ ⋮ $2n+5$
$→29$ ⋮ $2n+5$
$→2n+5∈Ư(29)=\{±1;±29\}$
$→n∈\{-2;-3;-17;12\}$