a, ĐKXĐ: \(x\ne\pm3\)

\(A=\frac{x\left(x-3\right)+2x\left(x+3\right)-3x^2-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)

\(=\frac{3x-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}=\frac{3x-12}{3x+9}\)

b, \(x=-4\Rightarrow A=\frac{3.\left(-4\right)-12}{3.\left(-4\right)+9}=8\)

c, \(A\in Z\Rightarrow3x-12⋮\left(3x+9\right)\Rightarrow3x+9-21⋮\left(3x+9\right)\Rightarrow21⋮\left(3x+9\right)\)

\(\Rightarrow3x+9\inƯ\left(21\right)=\left\{\pm1;\pm3;\pm7;\pm21\right\}\)

Mà \(3x+9⋮3\Rightarrow3x+9\in\left\{-21;-3;3;21\right\}\Rightarrow x\in\left\{-10;-4;-2;4\right\}\) (thỏa mãn điều kiện)

a, ĐỂ A xác định : 

\(\Rightarrow\hept{\begin{cases}x+3\ne0\\x-3\ne0\\x^2-9\ne0\end{cases}}\Rightarrow x\ne\pm3.\)

\(A=\left(\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x+3\right)\left(x-3\right)}\right):\frac{3}{x-3}\)

\(A=\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}:\frac{3}{x-3}\)

\(A=\frac{x^2-3x+2x^2+6x-3x^2+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)

\(A=\frac{3x+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)

\(A=\frac{x-4}{x+3}\)

b

Bài 1:

a: ĐKXĐ: x<>0; x<>3; x<>1

\(A=\left(\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\right):\frac{2x-2}{x}\)

\(=\frac{\left(x-3\right)^2-x^2+9}{x\left(x-3\right)}:\frac{2\left(x-1\right)}{x}\)

\(=\frac{x^2-6x+9-x^2+9}{x\cdot\left(x-3\right)}:\frac{2\left(x-1\right)}{x}=\frac{-6\left(x-3\right)}{x\left(x-3\right)}:\frac{2\left(x-1\right)}{x}=\frac{-6}{x}\cdot\frac{x}{2\left(x-1\right)}=\frac{-3}{x-1}\)

b: Để A là số nguyên thì -3⋮x-1

=>x-1∈{1;-1;3;-3}

=>x∈{2;0;4;-2}

Kết hợp ĐKXĐ, ta được: x∈{2;4;-2}

Bài 2:

a: \(x^3-2x^2=x^2\cdot x-x^2\cdot2=x^2\left(x-2\right)\)

b: \(y^2-2y-x^2+1\)

\(=\left(y^2-2y+1\right)-x^2\)

\(=\left(y-1\right)^2-x^2=\left(y-1-x\right)\left(y-1+x\right)\)

c: \(\left(x+1\right)^2-25\)

=(x+1+5)(x+1-5)

=(x-4)(x+6)

\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)

\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)

\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)

\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)

2x+3 chia hết cho 3x+1

=>3(2x+3) chia hết cho 3x+1

=> 6x+9 chia hết cho 3x+1

=>2(3x+1)+7 chia hết cho 3x+1 

=>7 chia hết cho 3x+1

=> 3x+1 thuộc Ư(7)=(1;7;-1;-7) 

=> x thuộc 0;2

a) bài 1

để \(x\in Z\)thì \(3x-1⋮x-1\)

mà \(x-1⋮x-1\)

\(\Rightarrow3\left(x-1\right)⋮x-1\)

\(\Rightarrow\left(3x-1\right)-\left[3x-3\right]⋮x-1\)

\(\Rightarrow2⋮x-1\)

\(\Rightarrow x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

ta có bảng

vậy \(x\in\left\{2;0;3;-1\right\}\)

còn nữa mà bạn