Mình bị nhầm:
Sửa lại thành (x+2y)(x - 2y - 2) = 3 nhé 
Rồi làm tương tự như dưới
Mình đang bận nên lát nữa trình bày lại sau!

Ta có:
\(x^2-4y^2-2x-4y=3\)
\(\Rightarrow x^2-\left(2y\right)^2-2\left(x-2y\right)=3\)
\(\Rightarrow\left(x-2y\right)\left(x+2y\right)-2\left(x-2y\right)=3\)
\(\Rightarrow\left(x-2y\right)\left(x+2y-2\right)=3\)
\(\Rightarrow x-2y;x+2y-2\in\)Ư(3)
Mà 3=3.1=1.3=-1.(-3)=-3.(-1)
Ta có bảng sau:
 

a) Ta có:
\(B=\dfrac{2x}{x^2-5x}\Rightarrow B=\dfrac{2x}{x\left(x-5\right)}\Rightarrow B=\dfrac{2}{x-5}\)
Mà \(A=\dfrac{2}{x-5}\)
\(\Rightarrow A=B\)

b) Ta có:
\(D=\dfrac{2x^2-3x-2}{3\left(2x+1\right)}\Rightarrow D=\dfrac{2x^2-4x+x-2}{3\left(2x+1\right)}\Rightarrow D=\dfrac{2x\left(x-2\right)+\left(x-2\right)}{3\left(2x+1\right)}\Rightarrow D=\dfrac{\left(2x+1\right)\left(x-2\right)}{3\left(2x+1\right)}\Rightarrow D=\dfrac{x-2}{3}\)
Mà \(C=\dfrac{x-2}{3}\)
\(\Rightarrow C=D\)

a)
\(\dfrac{6x^2+9x}{4x^2-9}=\dfrac{3x}{A}\Rightarrow A=3x.\dfrac{4x^2-9}{6x^2+9x}\Rightarrow A=3x.\dfrac{\left(2x\right)^2-3^2}{3x\left(2x+3\right)}\Rightarrow A=\dfrac{\left(2x-3\right)\left(2x+3\right)}{2x+3}\Rightarrow A=2x-3\)

b)

\(\dfrac{5\left(x+y\right)}{3}=\dfrac{5x^2-5y^2}{A}\Rightarrow A=5x^2-5y^2.\dfrac{3}{5\left(x+y\right)}\Rightarrow A=5\left(x^2-y^2\right).\dfrac{3}{5\left(x+y\right)}\Rightarrow A=\dfrac{3\left(x^2-y^2\right)}{x+y}\Rightarrow A=\dfrac{3\left(x-y\right)\left(x+y\right)}{x+y}\Rightarrow A=3\left(x-y\right)\)