Pthđgđ cua (C) va (d) la :
x3-(m-1)
x2+x+2m+1 =x+m+1 (*)
⇔x3-(m+1)x2+m=0
⇔(x-1)(x2-mx-m)=0
Đe (*) co 3 nghiem phan biet thi pt:x2-mx-m=0 phai co 2 nghiem phan biet⇔x⩽0 hoac x\geqslant4
goi x_{1},x_{2} la 2 nghiem cua pt tren \Rightarrowx_{1}+x_{2}=m va x_{1}x_{2}= -m
Ta co :f^{'}=3x^{2}-2(m+1)x +1
f^{'}(1)=2-2m
f^{'}(x_{1})=3x_{1}^{2}- 2(m+1)x +1
f^{'}(x_{2})=3x_{2}^{2}- 2(m+1)x+1
Theo đe bai ta co:f^{'}(1)+f^{'}(x_{1})+f^{'}(x_{2})=12
\Leftrightarrow2-2m+3(x_{1}^{2}+x_{2}^{2})-2(m+1)(x_{1}+x_{2})+2=12
\Leftrightarrow 2-2m+3[(x_{1}+x_{2})^{2}-2x_{1}x_{2}]-2(m+1)(x_{1}+x_{2})+2=12
Thay x_{1}+x_{2} va x_{1}x_{2}vao ta co pt:5m^{2}+6m-8=0 \Leftrightarrowm= -2(N) hoac m=\frac{4}{5}(L)
Vay voi m= -2 thi thoa ycđb