Ta có:\(\frac{x+9}{x-2}=\frac{x-2+11}{x-2}=1+\frac{11}{x-2}\)

Để \(\frac{x+9}{x-2}\in Z\) thì \(\frac{11}{x-2}\in Z\)

=>11 chia hết cho x-2

=>x-2\(\in\)Ư(11)={-11,-1,1,11}

=>x\(\in\){-9,1,3,13}

Để x+9 / x-2 thuộc Z thì tử chia hết cho mẫu

Tacos x+9 : x-2

=>x+9-(x-2) : x-2

11 : x-2

x-2 thuộc Ư(11)={-1;1;-11;1}

x thuộc {1;3;-9;3}

Vậy....

Mk chỉ lm mẫu cho bn 2 câu thôi , các câu khác tương tự nhóa ~~~

a, 10 chia hết cho n - 1

=> n - 1 thuộc Ư(10)

Mà : Ư(10) = { 1;2;5;10 }

+) n - 1 = 1 => n = 1 + 1 => n = 2

+) n - 1 = 2 => n = 2 + 1 => n = 3

+) n - 1 = 5 => n = 5 + 1 => n = 6

+) n - 1 = 10 => n = 10 + 1 => n = 11

Vậy n thuộc { 2;3;6;11 }

b, n + 9 chia hết cho n - 1

Mà : n - 1 chia hết cho n - 1

Nên : ( n + 9 ) - ( n - 1 ) chia hết cho n - 1

=> n + 9 - n + 1 chia hết cho n - 1

=> 10 chia hết cho n - 1 

=> n - 1 thuộc Ư(10)

Mà : Ư(10) = { 1;2;5;10 }

+) n - 1 = 1 => n = 1 + 1 =>n = 2

+) n - 1 = 2 =>n = 2 + 1 => n = 3

+) n - 1 = 5 => n = 5 + 1 => n = 6

+) n - 1 = 10 => n = 10 + 1 => n = 11

Vậy n thuộc { 2;3;6;11 }

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$\frac{1}{}$

$\frac{n+3}{2n-2}=\frac{n-4+7}{2n-2}=\frac{2(n-2)+7}{2n-2}=2+\frac{7}{2n-2}$

Để $\frac{n+3}{2n-2}$ thì $7⋮2n-2$

$\Rightarrow 2n-2\in Ư\left(7\right)\in \{\pm 1;\pm 7\}$

$2n-2=1\Rightarrow n=1,5$

$2n-2=-1\Rightarrow n=0,5$

$2n-2=7\Rightarrow n=4,5$

$2n-2=-7\Rightarrow n=-2,5$

Vì $n\in Z\Rightarrow$ Không có giá trị n thõa mãn