Đáp án:
a) \(m<0\)
c) \(y=24x-43\)
Giải thích các bước giải:
$$\eqalign{
& y = {x^4} + m{x^2} - m - 5 \cr
& y' = 4{x^3} + 2mx = 0 \Leftrightarrow \left[ \matrix{
x = 0 \hfill \cr
{x^2} = - {m \over 2} \hfill \cr} \right. \cr
& a)\,\,De\,\,do\,\,thi\,\,ham\,\,so\,\,co\,\,3\,\,CT \cr
& \Rightarrow y' = 0\,\,co\,\,3\,\,nghiem\,\,pb \cr
& \Rightarrow - {m \over 2} > 0 \Leftrightarrow m < 0 \cr
& b)\,\,Khi\,\,m = - 2 \Rightarrow y = {x^4} - 2{x^2} - 3 \cr
& y' = 4{x^3} - 4x = 0 \Leftrightarrow \left[ \matrix{
x = 0 \Rightarrow y = - 3 \hfill \cr
x = 1 \Rightarrow y = - 4 \hfill \cr
x = - 1 \Rightarrow y = - 4 \hfill \cr} \right. \cr
& Ban\,\,tu\,\,khao\,\,sat\,\,va\,\,ve\,\,nha!!! \cr
& c)\,\,Tiep\,\,tuyen\,\,//\,\,y = 24x + 1 \cr
& \Rightarrow y'\left( {{x_0}} \right) = 24 \Leftrightarrow 4{x^3} - 4x = 24 \cr
& \Leftrightarrow x = 2 \Rightarrow y = 5 \cr
& \Rightarrow pttt:\,\,y = 24\left( {x - 2} \right) + 5 = 24x - 43 \cr} $$
