Ta có \(f\left(x\right)>0,\forall x\in\left(0;1\right)\)

\(\Leftrightarrow-x^2-2\left(m-1\right)x+2m-1>0,\forall x\left(0;1\right)\)

\(\Leftrightarrow-2m\left(x-1\right)>x^2-2x+1,\forall x\in\left(0;1\right)\) (*)

Vì \(x\in\left(0;1\right)\Rightarrow x-1< 0\) nên (*) \(\Leftrightarrow-2m< \dfrac{x^2-2x+1}{x-1}=x-1=g\left(x\right),\forall x\in\left(0;1\right)\)

\(\Leftrightarrow-2m\le g\left(0\right)=-1\Leftrightarrow m\ge\dfrac{1}{2}\)

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: \(-2< m< 2\)

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

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

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

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

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

\(\Rightarrow m\ge1\)

\(x^2-5x+7+2m=0\Leftrightarrow x^2-5x+7=-2m\)

Xét hàm \(f\left(x\right)=x^2-5x+7\) trên \(\left[1;5\right]\)

\(-\dfrac{b}{2a}=\dfrac{5}{2}\in\left[1;5\right]\)

\(f\left(1\right)=3\) ; \(f\left(\dfrac{5}{2}\right)=\dfrac{3}{4}\) ; \(f\left(5\right)=7\)

\(\Rightarrow\) Pt đã cho có 2 nghiệm pb thuộc đoạn đã cho khi và chỉ khi:

\(\dfrac{3}{4}< -2m\le3\)

\(\Leftrightarrow-\dfrac{3}{2}\le m< \dfrac{3}{8}\)

Cả 4 đáp án đều sai là sao ta?

\(\frac{x^2+5x+a}{2x^2-3x+2}\ge-1\Leftrightarrow\frac{x^2+5x+a}{2x^2-3x+2}+1\ge0\Leftrightarrow\frac{3x^2+2x+a+2}{2x^2-3x+2}\ge0\)

\(\Leftrightarrow3x^2+2x+a+2\ge0\) \(\forall x\) (do \(2x^2-3x+2=2\left(x-\frac{3}{4}\right)^2+\frac{7}{8}>0\))

\(\Rightarrow\Delta'=1-3\left(a+2\right)=-5-3a\le0\Rightarrow a\ge\frac{-5}{3}\) (1)

Lại có: \(\frac{x^2+5x+a}{2x^2-3x+2}\le7\Leftrightarrow\frac{x^2+5x+a}{2x^2-3x+2}-7\le0\Leftrightarrow\frac{-13x^2+26x+a-14}{2x^2-3x+2}\le0\)

\(\Leftrightarrow-13x^2+26x+a-14\le0\) \(\forall x\)

\(\Rightarrow\Delta'=169+13\left(a-14\right)\le0\Rightarrow a\le-1\) (2)

Kết hợp (1) và (2) ta được: \(\frac{-5}{3}\le a\le-1\)

Đk: \(x\in R\)

Có \(2x^2-3x+2>0;\forall x\)

\(-1\le\dfrac{x^2+5x+m}{2x^2-3x+2}< 7\) với \(\forall x\)\(\Leftrightarrow-2x^2+3x-2\le x^2+5x+m< 14x^2-21x+14\) với mọi x

\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+2x+m+2\ge0;\forall x\left(1\right)\\13x^2-26x+14-m>0;\forall x\left(2\right)\end{matrix}\right.\)

Từ \(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a=3>0\left(lđ\right)\\\Delta\le0\end{matrix}\right.\)\(\Leftrightarrow4-4.3\left(m+2\right)\le0\)\(\Leftrightarrow-20-12m\le0\)\(\Leftrightarrow m\ge\dfrac{-5}{3}\)

Từ \(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}a=13>0\left(lđ\right)\\\Delta< 0\end{matrix}\right.\)\(\Leftrightarrow m< 1\)

Vậy \(-\dfrac{5}{3}\le m< 1\)