Giải thích các bước giải: muốn tìm đa thức A ta chỉ việc nhân chéo chia ngang 

VD:

a)Ta có:

$\frac{A}{2x-1}$ = $\frac{6x²+3x}{4x²-1}$ 

⇒A= $\frac{(6x²+3x)(2x-1)}{4x²-1}$ 

=$\frac{3x(2x+1)(2x-1)}{(2x-1)(2x+1)}$ = 3x

Vậy A=3x

b) $\frac{4x²-3x-7}{A}$=$\frac{4x-7}{2x+3}$   

⇒ A =$\frac{(4x²-3x-7)(2x+3)}{4x-7}$ 

=$\frac{(4x²+4x-7x-7)(2x+3)}{4x-7}$ 

=$\frac{[(4x²+4x]+[(-7x-7)](2x+3)}{4x-7}$ 

=$\frac{4x(x+1)-7(x+1)(2x+3)}{4x-7}$ 

=$\frac{(x+1)(4x-7)(2x+3)}{4x-7}$ 

=(x+1)(2x+3)

Vậy A=(x+1)(2x+3)

$\begin{array}{l}\quad \text{Hai phân thức bằng nhau:}\\ \quad \dfrac{A}{B} = \dfrac{C}{D} \Leftrightarrow A.D = B.C\\ a) \quad \dfrac{A}{2x - 1} = \dfrac{6x^2 + 3x}{4x^2 - 1}\\ \Leftrightarrow A = \dfrac{(6x^2 + 3x)(2x -1)}{4x^2 - 1}\\ \Leftrightarrow A = \dfrac{3x(2x + 1)(2x - 1)}{(2x-1)(2x+1)}\\ \Leftrightarrow A = 3x\\ b) \quad \dfrac{4x^2 - 3x - 7}{A} = \dfrac{4x-7}{2x+3}\\ \Leftrightarrow A = \dfrac{(4x^2 - 3x - 7)(2x + 3)}{4x - 7}\\ \Leftrightarrow A = \dfrac{(4x - 7)(x + 1)(2x +3)}{4x - 7}\\ \Leftrightarrow A = (x+1)(2x+3)\\ c)\quad \dfrac{4x^2 - 7x + 3}{x^2 -1} = \dfrac{A}{x^2 + 2x + 1}\\ \Leftrightarrow A = \dfrac{(4x^2 - 7x + 3)(x^2 + 2x + 1)}{x^2 - 1}\\ \Leftrightarrow A = \dfrac{(4x - 3)(x -1)(x+ 1)^2}{(x-1)(x+1)}\\ \Leftrightarrow A = (4x-3)(x+1)\\ d)\quad \dfrac{x^2 - 2x}{2x^2 -3x - 2} = \dfrac{x^2 + 2x}{A}\\ \Leftrightarrow A = \dfrac{(2x^2 - 3x - 2)(x^2 + 2x)}{x^2 - 2x}\\ \Leftrightarrow A = \dfrac{(2x+1)(x-2)(x+2)x}{x(x-2)}\\ \Leftrightarrow A = (2x+1)(x+2) \end{array}$