a, \(ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)

b, \(R=\left(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}+\frac{1}{x-1}\right):\frac{x^2+x}{x^3+x}\)

\(=\left(\frac{x^2-2x+1}{x^2+x+1}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right):\frac{x\left(x+1\right)}{x\left(x^2+1\right)}\)

\(=\left(\frac{\left(x^2-2x+1\right)\left(x-1\right)-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\right)\)

\(=\frac{x^3-3x^2+3x-1+3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)

\(=\frac{x^3-1}{x^3-1}.\frac{x^2+1}{x+1}=\frac{x^2+1}{x+1}\)

\(b,\) Để R = 0

\(\Leftrightarrow\frac{x^2+1}{x+1}=0\Leftrightarrow x^2+1=0\) ( vô lý)

Vậy ko có giá trị nào của x để R =0

\(c,\left|R\right|=1\Leftrightarrow\left[{}\begin{matrix}R=-1\\R=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{x^2+1}{x+1}=-1\\\frac{x^2+1}{x+1}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+1=-x-1\\x^2+1=x+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=0\\x^2-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Chưa đủ đề bạn ơi

1.

Nếu \(m=0\), \(f\left(x\right)=2x\)

\(\Rightarrow m=0\) không thỏa mãn

Nếu \(x\ne0\)

Yêu cầu bài toán thỏa mãn khi \(\left\{{}\begin{matrix}m< 0\\\Delta'=\left(m-1\right)^2-4m^2< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 0\\\left[{}\begin{matrix}m>1\\m< -\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow m< -\dfrac{1}{3}\)

2.

\(\dfrac{-x^2+2x-5}{x^2-mx+1}\le0\forall x\)

\(\Leftrightarrow\dfrac{-\left(x-1\right)^2-4}{x^2-mx+1}\le0\forall x\)

\(\Leftrightarrow x^2-mx+1>0\forall x\)

\(\Leftrightarrow\Delta=m^2-4< 0\Leftrightarrow-2< m< 2\)

Kết luận: \(-2< m< 2\)

Câu 2: 

a: Để f(x) chia hết cho g(x) thì \(2x^3+3x^2-x+4⋮2x+1\)

\(\Leftrightarrow2x^3+x^2+2x^2+x-2x-1+5⋮2x+1\)

\(\Leftrightarrow2x+1\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{0;-1;2;-3\right\}\)

b: Để f(x) chia hết cho g(x) thì \(3x^3-x^2+6x⋮3x-1\)

\(\Leftrightarrow3x^3-x^2+6x-2+2⋮3x-1\)

\(\Leftrightarrow3x-1\in\left\{1;-1;2;-2\right\}\)

hay \(x\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3}\right\}\)

Đặt \(x^2+4x+3=t\left(t\ge-1\right)\)

\(\left(x^2+4x+3\right)\left(x^2+4x+6\right)\ge m,\forall x\in R\)

\(\Leftrightarrow\left(x^2+4x+3\right)^2+3\left(x^2+4x+3\right)\ge m,\forall x\in R\)

\(\Leftrightarrow m\le f\left(t\right)=t^2+3t,\forall x\in R\)

Yêu cầu bài toán thỏa mãn khi:

\(m\le minf\left(t\right)=-2\)

Ta có: \(F\left(x\right)+G\left(x\right)-H\left(x\right)=0\)

\(\Leftrightarrow4x^2+3x-2+3x^2-2x+5-5x^2+2x-3=0\\ \Leftrightarrow2x^2+3x=0\\ \Rightarrow x\left(2x+3\right)=0\\ \Rightarrow x=0;x=\dfrac{-3}{2}\)

Vậy tìm được x thỏa mãn là: \(x=0;x=\dfrac{-3}{2}\)

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.

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

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

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

\(f'\left(x\right)=3\left(m-1\right)x^2+4\left(m-1\right)x+m\)

- Với \(m=1\Rightarrow f'\left(x\right)=1>0\) (không thỏa mãn)

- Với \(m\ne1\Rightarrow f'\left(x\right)< 0;\forall x\) khi và chỉ khi:

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

\(\Leftrightarrow\left\{{}\begin{matrix}1< m< 4\\m< 1\end{matrix}\right.\)

\(\Rightarrow\) Không tồn tại m thỏa mãn yêu cầu