\(\dfrac{n}{n-5}=\dfrac{n-5+5}{n-5}=\dfrac{n-5}{n-5}+\dfrac{5}{n-5}=1+\dfrac{5}{n-5}\)

Để \(\dfrac{n}{n-5}\in Z\) thì \(\dfrac{5}{n-5}\in Z\)

\(\Rightarrow n-5\in\text{Ư}_{\left(5\right)}=\left\{-5;-1;1;5\right\}\)

\(\Rightarrow n\in\left\{0;4;6;10\right\}\)

Do \(n\in N\) nên tất cả các giá trị đều nhận

Vậy ...

`(x-3/8) +1/4= 7/8`

`=> x-3/8 = 7/8-1/4`

`=> x-3/8 = 7/8-2/8`

`=> x-3/8 =5/8`

`=>x= 5/8 +3/8`

`=>x= 8/8=1`

khó thế =))

\(S=\dfrac{2^2}{3x5}+\dfrac{2^2}{5x7}+\dfrac{2^2}{7x9}+...+\dfrac{2^2}{97x99}\)

\(\dfrac{S}{2}=\dfrac{2}{3x5}+\dfrac{2}{5x7}+\dfrac{2}{7x9}+...+\dfrac{2}{97x99}\)

\(\dfrac{S}{2}=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}...+\dfrac{1}{97}-\dfrac{1}{99}=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)

S=\(\dfrac{64}{99}\)

Ta có : \(A\text{=}\dfrac{2023^{2023}}{2023^{2024}}\text{=}\dfrac{1}{2023}\)

và \(B\text{=}\dfrac{2023^{2022}}{2023^{2023}}\text{=}\dfrac{1}{2023}\)

\(\Rightarrow A\text{=}B\)

Ta có :

A=\(\dfrac{2023^{2023}}{2023^{2024}}\)=\(\dfrac{2023^{2022}.2023}{2023^{2023}.2023}\)=\(\dfrac{2023^{2022}}{2023^{2023}}\)

Mà B=\(\dfrac{2023^{2023}}{2023^{2024}}\)

Vậy A=B

ai giúp mik với

\(\left(-\dfrac{1}{6}\right)\cdot\dfrac{19}{4}\)
\(=-\dfrac{19}{24}\)

-1/6 . 19/4

=-2/12 . 57/12

=-114/12 = -19/2

chúc bạn học tốt
 

bằng 239/ 210 nha

\(\left(-\dfrac{1}{15}\right)-\dfrac{15}{14}\)
\(=-\dfrac{14}{210}-\dfrac{225}{210}\)
\(=-\dfrac{239}{210}\)