2 câu trả lời
$$\eqalign{ & y = {\left( {{x^2} - x} \right)^2}\,\,\left( {TXD:\,\,D = R} \right) \cr & y' = 2\left( {{x^2} - x} \right)\left( {2x - 1} \right) \cr & Cho\,\,y' = 0 \Leftrightarrow \left[ \matrix{ {x^2} - x = 0 \hfill \cr 2x - 1 = 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = 0 \hfill \cr x = 1 \hfill \cr x = {1 \over 2} \hfill \cr} \right. \cr & BXD: \cr & x\,\,\,\,\, - \infty \,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,{1 \over 2}\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\, + \infty \cr & y'\,\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,0\,\,\,\, + \,\,\,\,0\,\,\,\,\,\,\, - \,\,\,\,0\,\,\,\,\,\,\, + \cr & \Rightarrow Ham\,\,so\,\,NB/\left( { - \infty ;0} \right);\,\,\left( {{1 \over 2};1} \right) \cr & \,\,\,\,Ham\,\,so\,\,DB/\left( {0;{1 \over 2}} \right);\left( {1; + \infty } \right) \cr} $$
Đáp án:
Giải thích các bước giải: y'=2*(x^2-x)*(2x-1)
y'=0 => x=1/2 x=1 X=0