Pthđgđ cua (C) va (d) la :$x^{3}$-(m-1)$x^{2}$+x+2m+1 =x+m+1 (*)
$\Leftrightarrow$$x^{3}$-(m+1)$x^{2}$+m=0
$\Leftrightarrow$(x-1)($x^{2}$-mx-m)=0
Đe (*) co 3 nghiem phan biet thi pt:$x^{2}$-mx-m=0 phai co 2 nghiem phan biet$\Leftrightarrow$x$\leqslant$0 hoac x$\geqslant$4
goi $x_{1}$,$x_{2}$ la 2 nghiem cua pt tren $\Rightarrow$$x_{1}$+$x_{2}$=m va $x_{1}x_{2}$= -m
Ta co :$f^{'}$=3$x^{2}$-2(m+1)x +1
$f^{'}$(1)=2-2m
$f^{'}$($x_{1}$)=3$x_{1}^{2}$- 2(m+1)x +1
$f^{'}$($x_{2}$)=3$x_{2}^{2}$- 2(m+1)x+1
Theo đe bai ta co:$f^{'}(1)+f^{'}(x_{1})+f^{'}(x_{2})$=12
$\Leftrightarrow$2-2m+3($x_{1}^{2}+x_{2}^{2}$)-2(m+1)($x_{1}+x_{2}$)+2=12
$\Leftrightarrow $2-2m+3$[(x_{1}+x_{2})^{2}$-2$x_{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 $\Leftrightarrow$m= -2(N) hoac m=$\frac{4}{5}$(L)
Vay voi m= -2 thi thoa ycđb