\(\frac{2a+9}{a+3}-\frac{5a+17}{a+3}-\frac{3a}{a+3}=\frac{2a+9-5a-17-3a}{a+3}=\frac{-6a-8}{a+3}=\frac{-6a-18+10}{a+3}=\frac{-6\left(a+3\right)+10}{a+3}\)

\(\frac{2a+9}{a+3}-\frac{5a+17}{a+3}-\frac{3a}{a+3}\) là số nguyên

<=> a + 3 thuộc Ư(10) = {-10 ; -5 ; -2 ; -1 ; 1 ; 2 ; 5 ; 10}

<=> a thuộc {-13 ; -8 ; -5 ; -4 ; -2 ; -1 ; 2 ; 7}

câu a)

\(\frac{2a+8}{5}-\frac{a}{5}=\frac{2a+8-a}{5}=\frac{a+8}{5}\)

Để \(\frac{a+8}{5}\in Z\)thì \(a+8\)phải là bội của 5

Suy ra \(a+8\in\left\{\pm1;\pm5\right\}\)

Suy ra \(a\in\left\{-7;-9;-3;-13\right\}\)

Hết 

Câu 2 tương tự nha

bạn làm hộ mink câu b được không đúng mình k cho

Ta có:

B = \(\frac{2a+9}{a+3}-\frac{5a+17}{a+3}-\frac{3a}{a+3}\)

B = \(\frac{\left(2a+9\right)-\left(5a+17\right)-3a}{a+3}\)

B = \(\frac{2a+9-5a-17-3a}{a+3}\)

B = \(\frac{-6a-8}{a+3}=\frac{-6\left(a+3\right)+10}{a+3}=-6+\frac{10}{a+3}\)

Để B \(\in\)Z <=> 10 \(⋮\)a + 3  <=> a + 3 \(\in\)Ư(10) = {1; -1; 2; -2; 5; -5; 10; -10}

Lập bảng : 

Vậy ...

\(B=\frac{2a+9}{a+3}-\frac{5a+17}{a+3}-\frac{3a}{a+3}\)

\(B=\frac{2a+9-5a-17-3a}{a+3}\)

\(B=\frac{-6a-8}{a+3}\inℤ\)

\(\Leftrightarrow-6a-8⋮a+3\)

\(\Rightarrow-6a-18+10⋮a+3\)

\(\Rightarrow-6\left(a+3\right)+10⋮a+3\)

\(\Rightarrow10⋮a+3\)

\(\Rightarrow a+3\in\left\{-1;1;-2;2;-5;5;-10;10\right\}\)

\(\Rightarrow a\in\left\{-4;-2;-5;-1;-8;2;-13;7\right\}\)

Đặt \(D=\frac{2a+9}{a+3}+\frac{5a+17}{a+3}-\frac{3a}{a+3}\)

\(=\frac{2a+9+5a+17-3a}{a+3}\)

\(=\frac{4a+26}{a+3}=\frac{4\left(a+3\right)+14}{a+3}=4+\frac{14}{a+3}\)

\(\Rightarrow14⋮a+3\)

\(\Rightarrow a+3\inƯ\left(14\right)\)

Đến đây làm nốt

Đặt \(A=\frac{2a+9}{a+3}+\frac{5a+17}{a+3}-\frac{3a}{a+3}\)

\(\Rightarrow A=\frac{\left(2a+9\right)+\left(5a+17\right)-3a}{a+3}=\frac{4a+26}{a+3}=\frac{4a+12+14}{a+3}\)

\(=\frac{4\left(a+3\right)+14}{a+3}=4+\frac{14}{a+3}\)

Vì \(4\inℤ\)\(\Rightarrow\)Để A nguyên thì \(14⋮\left(a+3\right)\)\(\Rightarrow a+3\inƯ\left(14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

\(\Rightarrow a\in\left\{-17;-10;-5;-4;-2;-1;4;11\right\}\)

Vậy \(a\in\left\{-17;-10;-5;-4;-2;-1;4;11\right\}\)

\(\frac{2a+9}{a+3}+\frac{5a+17}{a+3}-\frac{3a}{a+3}=\frac{4a+26}{a+3}\)

Để Phân số trên nguyên

=> 4a + 26 chia hết cho a + 3

=> 4a + 12 + 14 chia hết cho a + 3

Vì 4a + 12 chia hết cho a + 3

=> 14 chia hết cho a + 3

=> a + 3 thuộc Ư(14)

=> a + 3 thuộc {1; -1; 2; -2; 7; -7; 14; -14}

=> a thuộc {-2; -4; -1; -5; 4; -11; 11; -17}

\(\frac{2a+9}{a+3}+\frac{5a+17}{a+3}-\frac{3a}{a+3}=\frac{2a+9+5a+17-3a}{a+3}=\frac{4a+26}{a+3}=\frac{4a+12+14}{a+3}\)

\(=\frac{4a+12}{a+3}+\frac{14}{a+3}=\frac{4\left(a+3\right)}{a+3}+\frac{14}{a+3}=4+\frac{14}{a+3}\in Z\)

\(\Rightarrow\frac{14}{a+3}\in Z\Rightarrow\)14 chia hết cho a+3

=>a+3=-14;-7;-2;-1;1;2;7;14

=>a=-17;-10;-5;-4;-2;-1;4;11

\(\frac{2a+9}{a+3}+\frac{5a+17}{a+3}-\frac{3a}{a+3}=\frac{2a+9+5a+17-3a}{a+3}=\frac{4a+26}{a+3}\)

=> 4a+26 chia het cho a+3

=> 4a+12+14 chia het cho a+3

=> 4(a+3) +14 chia het cho a+3

=> 14 chia het cho a+3

=> a+3= -1;1;-2;2;-7;7;-14;14

=> a= -4;-2;-5;-1;-10;4;-17;11