a: \(y=-x^3-\left(m+1\right)x^2+3\left(m+1\right)x\)

=>\(y'=-3x^2-\left(m+1\right)\cdot2x+3\left(m+1\right)\)

=>\(y'=-3x^2+x\cdot\left(-2m-2\right)+\left(3m+3\right)\)

Để hàm số nghịch biến trên R thì \(y'< =0\forall x\)

=>\(\left\{{}\begin{matrix}\text{Δ}< =0\\a< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(-2m-2\right)^2-4\cdot\left(-3\right)\left(3m+3\right)< =0\\-3< 0\end{matrix}\right.\)

=>\(4m^2+8m+4+12\left(3m+3\right)< =0\)

=>\(4m^2+8m+4+36m+36< =0\)

=>\(4m^2+44m+40< =0\)

=>\(m^2+11m+10< =0\)

=>\(\left(m+1\right)\left(m+10\right)< =0\)

TH1: \(\left\{{}\begin{matrix}m+1>=0\\m+10< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m>=-1\\m< =-10\end{matrix}\right.\)

=>\(m\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}m+1< =0\\m+10>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m< =-1\\m>=-10\end{matrix}\right.\)

=>-10<=m<=-1

b: \(y=-\dfrac{1}{3}x^3+mx^2-\left(2m+3\right)x\)

=>\(y'=-\dfrac{1}{3}\cdot3x^2+m\cdot2x-\left(2m+3\right)\)

=>\(y'=-x^2+2m\cdot x-\left(2m+3\right)\)

Để hàm số nghịch biến trên R thì \(y'< =0\forall x\)

=>\(\left\{{}\begin{matrix}\text{Δ}< =0\\a< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-1< 0\\\left(2m\right)^2-4\cdot\left(-1\right)\cdot\left(-2m-3\right)< =0\end{matrix}\right.\)

=>\(4m^2+4\left(-2m-3\right)< =0\)

=>\(m^2-2m-3< =0\)

=>(m-3)(m+1)<=0

TH1: \(\left\{{}\begin{matrix}m-3>=0\\m+1< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m>=3\\m< =-1\end{matrix}\right.\)

=>\(m\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}m-3< =0\\m+1>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m< =3\\m>=-1\end{matrix}\right.\)

=>-1<=m<=3

a.

\(\Leftrightarrow x^2+2\left(m-1\right)x+m^2+3m+5\ne0\) ; \(\forall x\)

\(\Leftrightarrow\Delta'=\left(m-1\right)^2-\left(m^2+3m+5\right)< 0\)

\(\Leftrightarrow-5m-4< 0\)

\(\Leftrightarrow m>-\dfrac{4}{5}\)

b. 

\(\Leftrightarrow x^2+2\left(m-1\right)x+m^2+m-6\ge0\) ;\(\forall x\)

\(\Leftrightarrow\Delta'=\left(m-1\right)^2-\left(m^2+m-6\right)\le0\)

\(\Leftrightarrow-3m+7\le0\)

\(\Rightarrow m\ge\dfrac{7}{3}\)

c.

\(x^2-2\left(m+3\right)x+m+9>0\) ;\(\forall x\)

\(\Leftrightarrow\Delta'=\left(m+3\right)^2-\left(m+9\right)< 0\)

\(\Leftrightarrow m^2+5m< 0\Rightarrow-5< m< 0\)

Để y xác định thì \(\left(m-2\right)x+2m-3\ge0\forall x\in\left[-1;4\right]\)

\(\Leftrightarrow mx-2x+2m-3\ge0\)

\(\Leftrightarrow m\left(x+2\right)-2x-3\ge0\)

\(\Leftrightarrow m\ge\dfrac{2x+3}{x+2}\left(x+2>0\forall x\in\left[-1;4\right]\right)\)

\(\Rightarrow1\le m\le\dfrac{11}{6}\)

Đề yêu cầu tìm m sao cho hàm số đồng biến trên từng khoảng xác định K \(\left(-\infty,m\right),\left(m,+\infty\right)\)

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

y đồng biến trên K \(\Leftrightarrow x^2-2mx+m^2-m+1\ge0,\forall x\in K\)

\(f\left(x\right)=x^2-2mx+m^2-m+1\ge0,\forall x\in K\) (1)

Nhận xét: f(x) là một parabol hướng lên và min tại \(x=m\)

(1) \(\Leftrightarrow\) \(f\left(m\right)\ge0\) \(\Leftrightarrow1\ge m\)

Vậy...

\(1.x^2+\dfrac{1}{x^2}-2m\left(x+\dfrac{1}{x}\right)+1+2m=0\left(1\right)\)\(đặt:x^2+\dfrac{1}{x^2}=t\)

\(x>0\Rightarrow t\ge2\sqrt{x^2.\dfrac{1}{x^2}}=2\)

\(x< 0\Rightarrow-t=-x^2+\dfrac{1}{\left(-x^2\right)}\ge2\Rightarrow t\le-2\)

\(\Rightarrow t\in(-\infty;-2]\cup[2;+\infty)\left(2\right)\)

\(\Rightarrow\left(1\right)\Leftrightarrow t^2-2mt+2m-1=0\)

\(\Leftrightarrow\left(t-1\right)\left(t-2m+1\right)=0\Leftrightarrow\left[{}\begin{matrix}t=1\notin\left(2\right)\\t=2m-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2m-1\le-2\\2m-1\ge2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}m\le-\dfrac{1}{2}\\m\ge\dfrac{3}{4}\end{matrix}\right.\)

\(2.\)  \(f^2\left(\left|x\right|\right)+\left(m-2\right)f\left(\left|x\right|\right)+m-3=0\left(1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}f\left(\left|x\right|\right)=-1\\f\left(\left|x\right|\right)=3-m\end{matrix}\right.\)

\(dựa\) \(vào\) \(đồ\) \(thị\) \(f\left(\left|x\right|\right)\) \(\Rightarrow f\left(\left|x\right|\right)=-1\) \(có\) \(2nghiem\) \(pb\)

\(\left(1\right)có\) \(6\) \(ngo\) \(pb\Leftrightarrow\left\{{}\begin{matrix}-1< 3-m< 3\\3-m\ne-1\\\end{matrix}\right.\)\(\Leftrightarrow0< m< 4\)

\(\Rightarrow m=\left\{1;2;3\right\}\)