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![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\Leftrightarrow\) \(4+\dfrac{8}{c-1}\in Z\) \(\Rightarrow\dfrac{8}{c-1}\in Z\) \(\Rightarrow8⋮\left(c-1\right)\) \(\Rightarrow c-1\in\left\{-8;-4;-2;-1;1;2;4;8\right\}\) \(\Rightarrow c\in\left\{-7;-3;-1;0;2;3;5;9\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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}
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\in Z\) \(\Leftrightarrow4+\dfrac{14}{n-4}\in Z\) \(\Rightarrow\dfrac{14}{n-4}\in Z\) \(\Rightarrow14⋮\left(n-4\right)\Rightarrow n-4\in\left\{-14;-7;-2;-1;1;2;7;14\right\}\)
\(\Rightarrow n\in\left\{-10;-3;2;3;5;6;11;18\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{8c+36}{c+7}=\frac{8c+56-20}{c+7}=\frac{8\left(c+7\right)}{c+7}-\frac{20}{c+7}=8-\frac{20}{c+7}\)
\(\Rightarrow\frac{8c+36}{c+7}\in Z\Leftrightarrow\frac{20}{c+7}\in Z\Leftrightarrow c+7\inƯ20\)
\(\Leftrightarrow c+7\in\left\{\pm1;\pm2;\pm4;\pm5;\pm10;\pm20\right\}\)
\(\Leftrightarrow c\in\left\{-27;-17;-12;-11;-9;-8;-6;-5;-3;-2;3;13\right\}\)
Vậy \(\Rightarrow\frac{8c+36}{c+7}\in Z\Leftrightarrow\frac{20}{c+7}\in Z\Leftrightarrow c+7\inƯ20\)
\(\Leftrightarrow c+7\in\left\{\pm1;\pm2;\pm4;\pm5;\pm10;\pm20\right\}\)
\(\Leftrightarrow c\in\left\{-27;-17;-12;-11;-9;-8;-6;-5;-3;-2;3;13\right\}\)
Vậy \(c\in\left\{-27;-17;-12;-11;-9;-8;-6;-5;-3;-2;3;13\right\}\) thì \(\frac{8c+36}{c+7}\) là số nguyên
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{8a-55}{a-5}\)có phải là phân số này không?
\(\frac{8a-55}{a-5}\)là số nguyên \(\Leftrightarrow8a-55⋮a-5\)
\(\Rightarrow8a-40-15⋮a-5\)
\(\Rightarrow8\left(a-5\right)-15⋮a-5\)
\(\Rightarrow15⋮a-5\)
\(\Rightarrow a-5\inƯ\left(15\right)=\left\{\pm1;\pm3;\pm5;\pm15\right\}\)
\(\Rightarrow a\in\left\{4;6;2;8;0;10;-10;20\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{5b+18}{b+6}\) ( A thuộc Z )
Ta có : \(A=\frac{5b+18}{b+6}=\frac{5b+30-12}{b+6}=5-\frac{12}{b+6}\)
Vì A thuộc Z nên 12 / b + 6 thuộc Z
\(\Rightarrow b+6\in\left\{\pm12;\pm6;\pm4;\pm3;\pm2;\pm1\right\}\)
\(\Rightarrow b\in\left\{-18;-12;-10;-9;-8;-7;-5;-4;-3;-2;0;6\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời :
Để \(\frac{2c+20}{c+7}\)nguyên
=> 2c + 20 \(⋮\)c + 7
=> 2 . (c + 14) + 6 \(⋮\)c + 7
=> 6 \(⋮\)c + 7
=> c + 7 \(\in\)Ư (6) = {1 ; - 1 ; 2 ; - 2 ; 3 ; - 3 ; 6 ; - 6}
=> c \(\in\){8 ; 6 ; 9 ; 5 ; 10 ; 4 ; 13 ; 1}