d: \(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{A}\)

hay A=x-2

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a. \(x^2y^3.35xy=5.7x^3y^4\)

\(\Leftrightarrow35x^3y^4=35x^3y^4\Rightarrowđpcm\)

\(b.x^2\left(x+2\right).\left(x+2\right)=x\left(x+2\right)^2.x\)

\(\Leftrightarrow x^2\left(x+2\right)^2=x^2\left(x+2\right)^2\Rightarrowđpcm\)

\(c.\left(3-x\right)\left(9-x^2\right)=\left(3+x\right)\left(x^2-6x+9\right)\)

\(\Leftrightarrow\left(3-x\right)\left(3-x\right)\left(3+x\right)=\left(3+x\right)\left(3-x\right)^2\)

\(\Leftrightarrow\left(3-x\right)^2\left(3+x\right)=\left(3-x\right)^2\left(3+x\right)\)

\(\Rightarrowđpcm\)

\(d.5\left(x^3-4x\right)=\left(10-5x\right)\left(-x^2-2x\right)\)

\(\Leftrightarrow5x^3-20x=5x^3-20x\Rightarrowđpcm\)

a) $\frac{3}{2x-3}$ và $\frac{3x+6}{2x^2+x-6}$

Cách 1: Dùng định nghĩa hai phân thức bằng nhau.

$\frac{3}{2x-3}$= $\frac{3x+6}{2x^2+x-6}$

Vì : $3\left(2x^2+x-6\right)=6x^2+3x-18$

=$6x^2+12x-9x-18$

=$2x\left(3x+6\right)-3\left(3x+6\right)$

=$\left(2x-3\right)\left(3x+6\right)$

Cách 2: Rút gọn phân thức

$\frac{3x+6}{2x^2+x-6}=\frac{}{}$

bài này mà ko bít làm

 ⇒ 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 

a)\(\frac{x^2+5x+4}{x^2-1}=\frac{A}{x^2-2x+1}\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+4\right)}{\left(x+1\right)\left(x-1\right)}=\frac{A}{\left(x-1\right)^2}\)

\(\Leftrightarrow\frac{x+4}{x-1}=\frac{A}{\left(x-1\right)^2}\). Nhân 2 vế ở tử với x-1 ta có:

\(x+4=\frac{A}{x-1}\Leftrightarrow A=\left(x-1\right)\left(x+4\right)=x^2+3x-4\)

b)\(\frac{x^2-3x}{2x^2-7x+3}=\frac{x^2+4x}{A}\)

\(\Leftrightarrow\frac{x\left(x-3\right)}{\left(2x-1\right)\left(x-3\right)}=\frac{x\left(x+4\right)}{A}\)

\(\Leftrightarrow\frac{x}{2x-1}=\frac{x\left(x+4\right)}{A}\).Nhân 2 vế ở mẫu với x ta có:

\(2x-1=\frac{x+4}{A}\)\(\Leftrightarrow\left(2x-1\right)\left(x+4\right)=A\Leftrightarrow A=2x^2+7x-4\)