A nguyên

=>10x-15+6 chia hết cho 2x-3

=>\(2x-3\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{2;1;3;0\right\}\)

Để C nguyên thì : 10x - 9 chia hết cho 2x - 3

<=> 10x - 15 + 6 chia hết cho 2x - 3

<=> 5(2x - 3) + 6 chia hết cho 2x - 3

=> 6 chia hết cho 2x - 3

=> 2x - 3 thuộc Ư(6) = {-6;-3;-2;-1;1;2;3;6}

Ta có bảng : 

Bài 1:

\(A=\frac{10x-9}{2x-3}=\frac{10x-15+6}{2x-3}=\frac{5.\left(2x-3\right)+6}{2x-3}=\frac{5.\left(2x-3\right)}{2x-3}+\frac{6}{2x-3}=5+\frac{6}{2x-3}\)

Để A nguyên thì \(\frac{6}{2x-3}\)nguyên

=> 6 chia hết cho 2x - 3

=> \(2x-3\inƯ\left(6\right)\)

Mà 2x - 3 là số lẻ => \(2x-3\in\left\{1;-1;3;-3\right\}\)

=> \(2x\in\left\{4;2;6;0\right\}\)

=> \(x\in\left\{2;1;3;0\right\}\)

Vậy \(x\in\left\{2;1;3;0\right\}\)thỏa mãn đề bài

Bài 2:

\(3+\frac{a}{b}=3.\frac{a}{b}\)

=> \(3.\frac{a}{b}-\frac{a}{b}=3\)

=> \(2.\frac{a}{b}=3\)

=> \(\frac{a}{b}=\frac{3}{2}\)

Vậy \(\frac{a}{b}=\frac{3}{2}\)

`M=(10x^2-7x-5)/(2x-3)(x ne 3/2)`

`=(10x^2-15x+8x-12+7)/(2x-3)`

`=(5x(2x-3)+4(2x-3)+7)/(2x-3)`

`=5x+4+7/(2x-3)`

Để `M in ZZ`

`=>7/(2x-3) in ZZ`

`=>2x-3 in Ư(7)={+-1,+-7}`

`=>2x in {2,4,-4,10}`

`=>x in {1,2,-2,5}(tm)`

Vậy `x in {1,2,-2,5}` thì `M in ZZ`.

a) ĐKXĐ: \(x\notin\left\{3;-3;-2\right\}\)

Ta có: \(P=\left(\dfrac{2x-1}{x+3}-\dfrac{x}{3-x}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\)

\(=\left(\dfrac{\left(2x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3-10x}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{x+2}{x-3}\)

\(=\dfrac{2x^2-6x-x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+2}{x-3}\)

\(=\dfrac{3x^2+6x}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+2}{x-3}\)

\(=\dfrac{3x\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x+2}\)

\(=\dfrac{3x}{x+3}\)

b) Ta có: \(x^2-7x+12=0\)

\(\Leftrightarrow x^2-3x-4x+12=0\)

\(\Leftrightarrow x\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=4\left(nhận\right)\end{matrix}\right.\)

Thay x=4 vào biểu thức \(P=\dfrac{3x}{x+3}\), ta được: 

\(P=\dfrac{3\cdot4}{4+3}=\dfrac{12}{7}\)

Vậy: Khi \(x^2-7x+12=0\) thì \(P=\dfrac{12}{7}\)

\(\text{Để }\frac{10x^2-7x-5}{2x-3}nguyên\Rightarrow\left(10x^2-7x-5\right)⋮\left(2x-3\right)\)

\(\text{Ta có }10x^2-7x-5=10x^2-7x-12+7=\left(2x-3\right)\left(5x+4\right)+7\)\(Mà\left(2x-3\right)\left(5x+4\right)⋮\left(2x-3\right)\Rightarrow7⋮\left(2x-3\right)\)

\(\Rightarrow\left(2x-3\right)\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

\(\text{Vậy x }\in\left\{-2;1;2;5\right\}\)

\(a,P=\left(\dfrac{2x-1}{x+3}-\dfrac{x}{3-x}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\left(x\ne\pm3;x\ne-2\right)\\ P=\dfrac{2x^2-7x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x+2}\\ P=\dfrac{3x^2+6x}{\left(x-3\right)\left(x+2\right)}=\dfrac{3x\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}=\dfrac{3x}{x-3}\\ b,x^2-7x+12=0\\ \Leftrightarrow\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow x=4\left(x\ne3\right)\\ \Leftrightarrow A=\dfrac{3\cdot4}{4-3}=12\\ c,P=\dfrac{3\left(x-3\right)+9}{x-3}=3+\dfrac{9}{x-3}\in Z\\ \Leftrightarrow x-3\inƯ\left(9\right)=\left\{-9;-3;-1;1;3;9\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;4;6;12\right\}\)

\(A=\frac{5x+9}{x+1}=\frac{5x+5+4}{x+1}\)\(ĐKXĐ:x\ne-1\)

\(=\frac{5x+5}{x+1}+\frac{4}{x+1}\)

\(=\frac{5\left(x+1\right)}{x+1}+\frac{4}{x+1}\)

\(=5+\frac{4}{x+1}\)

\(\Rightarrow A=5+\frac{4}{x+1}\)

Để \(A\in Z\Rightarrow5+\frac{4}{x+1}\in Z\)

\(\Rightarrow x+1\inƯ\left(4\right)=\left\{1;2;4;-1;-2;-4\right\}\)

\(\Rightarrow x=\left\{0;1;3;-2;-3;-5\right\}\)