1.
ĐK: $A\ne 0; x\ne -1,5$
$\dfrac{4x^2-3x-7}{A}=\dfrac{4x-7}{2x+3}$
$\Leftrightarrow A=\dfrac{(2x+3)(4x^2-3x-7)}{4x-7}$
$=\dfrac{(2x+3)(4x-7)(x+1)}{4x-7}$
$=(2x+3)(x+1)$
$=2x^2+5x+3$
2.
$\dfrac{x^2+x-2}{x^2}=\dfrac{(x-1)(x+2)}{x^2}$
$\dfrac{x+2}{x+1}$
$\dfrac{x^2-4}{x^2-x-2}=\dfrac{(x-2)(x+2)}{(x+1)(x-2)}=\dfrac{x+2}{x+2}$
Vậy phân thức 2 bằng phân thức 3.