Đáp án:
$12)y=\dfrac{x^2-x+1}{x-1}$ $⇒y'=(\dfrac{x^2-x+1}{x-1})'$$=(\dfrac{x^2-x+1}{0x^2+x-1})'$
$=\dfrac{[1.1-(-1).0]x^2+2[1.(-1)+1.0]x+[(-1).(-1)-1.1]}{(x-1)^2}$
$=\dfrac{x^2-2x}{(x-1)^2}$
Vậy $a=1;b=-2⇒a.b=1.(-2)=-2$
$13) f(x)= -2x^4+4x^2+1$
$⇒f'(x)=(-2x^4+4x^2+1)'$
$=-2.4x^3+4.2x=-8x^3+8x$
Để $f'(x)<0⇒-8x^3+8x<0=8x(-x^2+1)⇔$\(\left[ \begin{array}{l}-1<x<0\\1<x\end{array} \right.\)
BẠN THAM KHẢO.
