Đáp án:
$a)y=\dfrac{6x^2-3x+2}{3x^2+2x-3}$$\dfrac{x}{y}$
$y'=\dfrac{(6x^2-3x+2)'(3x^2+2x-3)-(6x^2-3x+2)(3x^2+2x-3)'}{(3x^2+2x-3)^2}$
$=\dfrac{(12x-3)(3x^2+2x-3)-(6x^2-3x+2).(6x+2)}{(3x^2+2x-3)^2}$
$=\dfrac{36x^3+24x^2-36x-9x^2-6x+9-(36x^3+12x^2-18x^2-6x+12x+4)}{(3x^2+2x-3)^2}$
$=\dfrac{36x^3+24x^2-36x-9x^2-6x+9-36x^3-12x^2+18x^2+6x-12x-4}{(3x^2+2x-3)^2}$
$=\dfrac{21x^2-48x+5}{(3x^2+2x-3)^2}$
$b)y=\dfrac{2x+1}{2x^2-3x+1}$
$y'=\dfrac{(2x+1)'(2x^2-3x+1)-(2x+1)(2x^2-3x+1)'}{(2x^2-3x+1)^2}$
$=\dfrac{2(2x^2-3x+1)-(2x+1)(4x-3)}{(2x^2-3x+1)^2}$
$=\dfrac{4x^2-6x+2-(8x^2-6x+4x-3)}{(2x^2-3x+1)^2}$
$=\dfrac{4x^2-6x+2-8x^2+6x-4x+3}{(2x^2-3x+1)^2}$
$=\dfrac{-4x^2-4x+5}{(2x^2-3x+1)^2}$
BẠN THAM KHẢO.
