Đáp án:
$13)y=\dfrac{2x^2-5x+6}{4-3x}$ $⇒y'=(\dfrac{2x^2-5x+6)}{4-3x})'$
$=(\dfrac{2x^2-5x+6}{0x^2-3x+4})'$
$=\dfrac{[2.(-3)-(-5).0]x^2+2[2.4-6.0]x+[(-5).4-6.(-3)]}{(4-3x)^2}$
$=\dfrac{-6x^2+16x-2}{(4-3x)^2}$
$14)y=\dfrac{5x^2-4x-9}{-2x^2+3x-8}$ $⇒y'=(\dfrac{5x^2-4x-9}{-2x^2+3x-8})'$
$=\dfrac{[5.3-(-4).(-2)]x^2+2[5.(-8)-(-9).(-2)x+[(-4).(-8)-(-9).3]}{(-2x^2+3x-8)^2}$
$=\dfrac{7x^2-116x+59}{(-2x^2+3x-8)^2}$
$15)y=\dfrac{x^3}{2-x}$ $⇒y'=\dfrac{x^3}{2-x}$
$=\dfrac{(x^3)'(2-x)+x^3.(2-x)'}{(2-x)^2}$
$=\dfrac{3x^2(2-x)-x^3.(-1)}{(2-x)^2}$
$=\dfrac{6x^2-3x^3+x^3}{(2-x)^2}$
$=\dfrac{-2x^3+6x^2}{(2-x)^2}$
BẠN THAM KHẢO.
