Đáp án:
$y=\dfrac{2x^2-3x+7}{x^2+2x+3}$
$y'=\dfrac{(2x^2-3x+7)'(x^2+2x+3)-(2x^2-3x+7)(x^2+2x+3)'}{(x^2+2x+3)^2}$
$=\dfrac{(4x-3)(x^2+2x+3)-(2x^2-3x+7)(2x+2)}{(x^2+2x+3)^2}$
$=\dfrac{4x^3+8x^2+12x-3x^2-6x-9-(4x^3+4x^2-6x^2-6x+14x+14)}{(x^2+2x+3)^2}$
$=\dfrac{4x^3+8x^2+12x-3x^2-6x-9-4x^3-4x^2+6x^2+6x-14x-14}{(x^2+2x+3)^2}$
$=\dfrac{7x^2-2x-23}{(x^2+2x+3)^2}$
BẠN THAM KHẢO.