\(y'=\dfrac{-m^2-1}{\left(x-m\right)^2}\)

\(y'< 0\) ;\(\forall x\in\left(0;1\right)\Leftrightarrow\left[{}\begin{matrix}m\ge1\\m\le0\end{matrix}\right.\)

\(y'=\dfrac{-2m-1}{\left(x-2\right)^2}\)

\(y'< 0\) với mọi x thuộc TXĐ \(\Leftrightarrow-2m-1< 0\Leftrightarrow m>-\dfrac{1}{2}\)

Đặt \(g\left(x\right)=\left(1+x\right)\left(2+x\right)...\left(2017+x\right)\)

\(\Rightarrow g\left(0\right)=1.2.3...2017=2017!\)

\(f\left(x\right)=\dfrac{x}{g\left(x\right)}\Rightarrow f'\left(x\right)=\dfrac{g\left(x\right)-x.g'\left(x\right)}{g^2\left(x\right)}\)

\(\Rightarrow f'\left(0\right)=\dfrac{g\left(0\right)-0.g'\left(x\right)}{\left[g\left(0\right)\right]^2}=\dfrac{g\left(0\right)}{\left[g\left(0\right)\right]^2}=\dfrac{1}{g\left(0\right)}=\dfrac{1}{2017!}\)

\(f'\left(x\right)=2ax+b\)

\(f\left(x\right)+\left(x-1\right)f'\left(x\right)=ax^2+bx+c+\left(x-1\right)\left(2ax+b\right)\)

\(=3ax^2+\left(2b-2a\right)x+c-b\)

Yêu cầu bài toán thỏa mãn khi: \(\left\{{}\begin{matrix}3a=3\\2b-2a=0\\c-b=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c=1\)

\(y'=x^2-2x+m\)

\(y'\ge0\) ; \(\forall x\in\left(1;3\right)\Leftrightarrow x^2-2x+m\ge0\) ;\(\forall x\in\left(1;3\right)\)

\(\Leftrightarrow m\ge\max\limits_{\left(1;3\right)}\left(-x^2+2x\right)\)

Xét hàm \(f\left(x\right)=-x^2+2x\) trên \(\left(1;3\right)\)

\(-\dfrac{b}{2a}=1\) ; \(f\left(1\right)=1\) ; \(f\left(3\right)=-3\)

\(\Rightarrow m\ge1\)

\(y'=3x^2-6x+3=3\left(x^2-2x+1\right)=3\left(x-1\right)^2\ge0\)

\("="\Leftrightarrow x=1\)

\(y'=4mx^3+2mx=2mx\left(2x^2+1\right)\)

Do \(2x\left(x^2+1\right)>0\) ;\(\forall x>0\)

\(\Rightarrow y'\ge0\) ;\(\forall x>0\) khi và chỉ khi \(m>0\)

\(y'=3x^2-2x+m\)

\(y'\ge0\Leftrightarrow\left\{{}\begin{matrix}a>0\\\Delta'\le0\end{matrix}\right.\Leftrightarrow1-3m\le0\Leftrightarrow m\ge\dfrac{1}{3}\)

\(y'=x^2-2mx+m\)

\(y'\ge0\Leftrightarrow\left\{{}\begin{matrix}a>0\\\Delta'\le0\end{matrix}\right.\Leftrightarrow m^2-m\le0\Leftrightarrow0\le m\le1\)