Hướng dẫn giải:

 ⇒ x 2 - 2 x . A = 2 x 2 - 3 x - 2 x 2 + 2 x

      ⇒ x x - 2 . A = 2 x 2 - 4 x + x - 2 . x x + 2

      ⇒  x x - 2 . A = 2 x x - 2 + x - 2 . x x + 2

      ⇒ x(x – 2).A = (x – 2)(2x + 1).x.(x + 2)

      ⇒ A = (2x + 1)(x + 2) = 2 x 2 + 4 x + x + 2 = 2 x 2 + 5 x + 2

Vậy 

 ⇒ 4 x 2 - 7 x + 3 x 2 + 2 x + 1 = A x 2 - 1

      ⇒ 4 x 2 - 4 x - 3 x + 3 x + 1 2 = A x + 1 x - 1

      ⇒ 4 x x - 1 - 3 x - 1 . x + 1 2 = A . x + 1 x - 1

      ⇒ x - 1 4 x - 3 x + 1 2 = A x + 1 x - 1

      ⇒ A = 4 x - 3 x + 1 = 4 x 2 + 4 x - 3 x - 3 = 4 x 2 + x - 3

Vậy 

\(\dfrac{A}{x-3}=\dfrac{y-x}{3-x}\)

\(\Rightarrow A=\dfrac{\left(x-3\right)\left(y-x\right)}{3-x}\)

\(\Rightarrow A=\dfrac{-\left(3-x\right)\left(y-x\right)}{3-x}\)

\(\Rightarrow A=x-y\)

_____

\(\dfrac{5x}{x+1}=\dfrac{Ax\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}\)

\(\Rightarrow A=\dfrac{5x\left(x+1\right)\left(1-x\right)}{x\left(x+1\right)}\)

\(\Rightarrow A=5\left(1-x\right)\)

\(\Rightarrow A=5-5x\)

____
\(\dfrac{4x^2-5x+1}{A}=\dfrac{4x-1}{x+3}\)

\(\Rightarrow\dfrac{\left(4x-1\right)\left(x-1\right)}{A}=\dfrac{4x-1}{x+3}\)

\(\Rightarrow A=\dfrac{\left(4x-1\right)\left(x-1\right)\left(x+3\right)}{4x-1}\)

\(\Rightarrow A=\left(x-1\right)\left(x+3\right)\)

\(\Rightarrow A=x^2+2x-3\)

 ⇒ 4 x 2 - 3 x - 7 2 x + 3 = A 4 x - 7

      ⇒ 4 x 2 + 4 x - 7 x - 7 2 x + 3 = A 4 x - 7

      ⇒ [4x(x + 1) – 7(x + 1)](2x+ 3) = A(4x - 7)

      ⇒ (x + 1)(4x – 7)(2x + 3) = A(4x – 7)

      ⇒ A = (x + 1)(2x + 3) = 2 x 3 + 3 x + 2 x + 3 = 2 x 2 + 5 x + 3

Vậy 

Bài 1: 

c: ĐKXĐ: \(x\notin\left\{-1;3\right\}\)