Ta có 

A = \(\frac{n-3}{2n-1}-\frac{n-5}{2n-1}\)

= \(\frac{(n-3)-(n-5)}{2n-1}\)

= \(\frac{n-3-n+5}{2n-1}\)

= \(\frac{n-n-3+5}{2n-1}\)

= \(\frac{2}{2n-1}\)

Để \(\frac{2}{2n-1}\inℕ\)

=> \(2⋮2n-1\)

=> \(2n-1\inƯ\left(2\right)\)

=> \(2n-1\in\left\{1;2\right\}\)

Xét từng trường hợp ta có : 

+) 2n - 1 = 1

=> 2n = 1 + 1

=> 2n = 2

=> n = 2 : 2

=> n = 1 (chọn)

+) 2n - 1 = 2

=> 2n = 2 + 1

=> 2n = 3

=> n = 3 : 2

=> n = 1,5 (loại)

Vậy n = 1 

\(A=\frac{n-3}{2n-1}-\frac{n-5}{2n-1}=\frac{\left(n-3\right)-\left(n-5\right)}{2n-1}=\frac{2}{2n-1}\)

Để \(A\in Z\)thì \(\frac{2}{2n-1}\in Z\)hay \(\left(2n-1\right)\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Mà \(n\in N\)

\(\Rightarrow n\in\left\{0;1;\frac{3}{2}\right\}\)

vì n thuộc N suy ra 2n là số chẵn

                 ______2n-1 là số lẻ

                 _______2n-1 thuộc {1;-1}

_2n-1=1suy ra n=1

_2n-1=-1suy ra n=0

  vậy n =1;0

a) \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)

\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)

\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\left(\frac{1}{2}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=\frac{-1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{-1}{6}\end{cases}}}\)

Vậy x= 5/6 hoặc -1/6

b) - Nếu x=0 thì \(5^y=2^0+624=1+624=625=5^4\Rightarrow y=4\left(y\in N\right)\)

- Nếu x \(\ne\) 0 thì vế trái là số chẵn , vế phải là số lẻ \(\forall x;y\inℕ\) ( vô lí)

Vậy x=0, y=4

thank you bạn.

2 mũ x ×3mũ g nha các bn mk ghi thiếu

\(a,\\ =>n-3\inƯ\left(-7\right)\\ Ư\left(-7\right)=\left\{1;-1;7;-7\right\}\\ =>\left\{{}\begin{matrix}n-3=1\\n-3=-1\\n-3=7\\n-3=-7\end{matrix}\right.\\ =>\left\{{}\begin{matrix}n=4\\n=2\\n=10\\n=-4\end{matrix}\right.\)

\(b,\dfrac{n-5}{n+1}=1-\dfrac{6}{n+1}\\ =>n+1\inƯ\left(6\right)\\ Ư\left(6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\\ =>\left\{{}\begin{matrix}\left\{{}\begin{matrix}n+1=1\\n+1=-1\\n+1=2\\n+1=-2\end{matrix}\right.\\\left\{{}\begin{matrix}n+1=3\\n+1=-3\\n+1=6\\n+1=-6\end{matrix}\right.\end{matrix}\right.=>\left\{{}\begin{matrix}\left\{{}\begin{matrix}n=0\\n=-2\\n=1\\n=-3\end{matrix}\right.\\\left\{{}\begin{matrix}n=2\\n=-4\\n=5\\n=-7\end{matrix}\right.\end{matrix}\right.\)

sao khongg ai giúp minhh :<

Ta có :

\(n^2+9n+9=n.\left(n+9\right)+9=n.\left(n-4\right)+13n+9\) chia hết cho n - 4

\(\Leftrightarrow13n+9=13n-52+61\) chia hết cho n - 4

\(\Leftrightarrow61\) chia hết cho n - 4

\(\Leftrightarrow n-4\inƯ\left(61\right)\)

\(\Leftrightarrow n-4\in\left\{1;61\right\}\)

\(\Leftrightarrow n\in\left\{5;65\right\}\)

!ERROR 404!

\(a^n=1\Rightarrow a^n=a^0\Rightarrow\left\{{}\begin{matrix}n=0\\a\in N\end{matrix}\right.\)