\(\text{Ta có:}\)

\(\text{Để}\)\(\frac{4b+42}{b+7}\)\(\text{nguyên thì}\)\(4b+42⋮b+7\)

\(\text{Lại có:}\)

\(\text{4b + 42 = 4b + 28 + 14 = 4( b+7 ) + 14}\)

\(\text{Vì}\)\(b+7⋮b+7\)\(\Rightarrow4\left(b+7\right)⋮b+7\)

\(\text{Do đó:}\)\(14⋮b+7\)

\(\Rightarrow b+7\inƯ\left(14\right)=\left\{1;2;7;14\right\}\)

\(\Rightarrow b\in\left\{-6;-5;0;7\right\}\)

b ∈{3,5,7} 

Vì \(\frac{15}{x}+4\) là số nguyên

    \(\Rightarrow15⋮x\)(hoặc \(x\inƯ\left(15\right)\)

 Vậy Ư(15)là:[1,-1,3,-3,5,-5,15,-15]

              Do đó \(x\in\)[1,-1,3,-3,5,-5,15,-15]

để phân số trên là số nguyên thì (x+4) thuộc Ư(15)={1,3,5,-1,-3,-5,15,-15}

xét từng TH:

x+4=1=>x=-3

x+4=3=>x=-1

x+4=5=>x=1

x+4=15=>x=11

x+4=-1=>x=-5

x+4=-3=>x=-7

x+4=-5=>x=-9

x+4=-15=>x=-19

vậy x thuộc { -19,-9,-7,-5,-1,1,11,-3}

Để \(\frac{15}{x+4}\) là số nguyên thì \(x+4\inƯ\left(15\right)=\left\{-15;-5;-3;-1;1;3;5;15\right\}\)

\(\Rightarrow x\in\left\{-19;-9;-7;-5;-3;-1;1;11\right\}\)

4x-37 chia hết cho x-6

4x-24-13

=>13 chia hết cho x-6

x=7,19,5,-7

Ta đặt A\(=\dfrac{4c-4+8}{c-1}\) \(\Rightarrow A=\dfrac{4c-4+8}{c-1}=\dfrac{4\left(c-1\right)+8}{c-1}=4+\dfrac{8}{c-1}\)

Để A∈Z \(\Leftrightarrow\) \(4+\dfrac{8}{c-1}\in Z\) \(\Rightarrow\dfrac{8}{c-1}\in Z\) \(\Rightarrow8⋮\left(c-1\right)\) \(\Rightarrow c-1\in\left\{-8;-4;-2;-1;1;2;4;8\right\}\) \(\Rightarrow c\in\left\{-7;-3;-1;0;2;3;5;9\right\}\)

Ta đặt A=\(\dfrac{4n-2}{n-4}\)\(\Rightarrow A=\dfrac{4n-16+14}{n-4}=\dfrac{4\left(n-4\right)+14}{n-4}=4+\dfrac{14}{n-4}\)

Để A\(\in Z\) \(\Leftrightarrow4+\dfrac{14}{n-4}\in Z\) \(\Rightarrow\dfrac{14}{n-4}\in Z\) \(\Rightarrow14⋮\left(n-4\right)\Rightarrow n-4\in\left\{-14;-7;-2;-1;1;2;7;14\right\}\) 

\(\Rightarrow n\in\left\{-10;-3;2;3;5;6;11;18\right\}\)

Giải:
Để \(\frac{6m-20}{m-5}\in Z\Rightarrow6m-20⋮m-5\)

\(\Rightarrow\left(6m-30\right)+10⋮m-5\)

\(\Rightarrow6\left(m-5\right)+10⋮m-5\)

\(\Rightarrow10⋮m-5\)

\(\Rightarrow m-5\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

\(\Rightarrow m\in\left\{6;4;7;3;10;0;15;-5\right\}\)

Vậy...