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

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

\(y'=-3x^2-6x+m\Rightarrow y''=-6x-6\)

\(y''=0\Leftrightarrow-6x-6=0\Leftrightarrow x=-1\notin\left[0;1\right]\)

\(\left\{{}\begin{matrix}y'\left(0\right)=m\\y'\left(1\right)=m-9\end{matrix}\right.\Rightarrow^{max}_{\left[0;1\right]}y'=y'\left(0\right)=m\)

\(\Rightarrow m=10\)

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

\(y'=0\Leftrightarrow mx^2+2\left(1-m\right)x-m=0\)

Xet \(m=0\) ko thoa man pt

Xet \(m\ne0\)

\(\left\{{}\begin{matrix}\Delta'>0\\\dfrac{2\left(m-1\right)}{m}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(1-m\right)^2+m^2>0\left(ld\right)\\m=-2\end{matrix}\right.\Rightarrow m=-2\)

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

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

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

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

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

\(\Rightarrow m\ge1\)

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

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

a/ \(y'=3mx^2-2\left(m+1\right)x+3m\)

Xet m=0 ko thoa man

Xet m khac 0

\(y'\ge0\Leftrightarrow\left(m+1\right)^2-9m^2\le0\Leftrightarrow8m^2-2m-1\ge0\)

\(\Leftrightarrow m^2+8\le0\left(vl\right)\) => ko ton tai m thoa man

b/ \(y'=mx^2-2mx+2m-1\)

m=0 ko thoa man

Xet m khac 0

\(y'\ge0\Leftrightarrow\left\{{}\begin{matrix}m>0\\m^2-m\left(2m-1\right)\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>0\\m^2-m\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>0\\\left[{}\begin{matrix}m\ge1\\m\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow m\ge1\)

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

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

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

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

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