Đáp án B.

a.

- Với \(m=-1\Rightarrow x=\dfrac{6}{7}\) (ktm)

- Với \(m\ne-1\) 

\(\Delta=\left(8m+1\right)^2-24m\left(m+1\right)=40m^2-8m+1>0;\forall m\) \(\Rightarrow\) pt luôn có 2 nghiệm pb

Để pt có 2 nghiệm thỏa mãn: \(x_1< x_2\le1\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x_1-1\right)\left(x_2-1\right)\ge0\\\dfrac{x_1+x_2}{2}< 1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_1x_2-\left(x_1+x_1\right)+1\ge0\\x_1+x_2< 2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6m}{m+1}-\dfrac{8m+1}{m+1}+1\ge0\\\dfrac{8m+1}{m+1}< 2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-m}{m+1}\ge0\\\dfrac{6m-1}{m+1}< 0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}-1< m\le0\\-1< m< \dfrac{1}{6}\end{matrix}\right.\) \(\Rightarrow-1< m\le0\)

\(\Rightarrow\) Pt có nghiệm thuộc khoảng đã cho khi: \(\left[{}\begin{matrix}m>0\\m< -1\end{matrix}\right.\)

b.

Đặt \(f\left(x\right)=\left(m+1\right)x^2-\left(8m+1\right)x+6m\)

Pt đã cho có đúng 1 nghiệm thuộc (0;1) khi:

\(f\left(0\right).f\left(1\right)< 0\)

\(\Leftrightarrow6m\left(m+1-8m-1+6m\right)< 0\)

\(\Leftrightarrow-6m^2< 0\)

\(\Leftrightarrow m\ne0\)

\(a=-1< 0;\Delta=\left(2\sqrt{m}-1\right)^2+4\left(\sqrt{m}-m\right)=4m-4\sqrt{m}+1+4\sqrt{m}-4m=1>0\)

a/ \(f\left(x\right)\ge0\) vô nghiệm \(\Leftrightarrow f\left(x\right)< 0,\forall x\in R\Leftrightarrow\left\{{}\begin{matrix}a=-1< 0\left(tm\right)\\\Delta< 0\left(voly\right)\end{matrix}\right.\)

Vậy ko tồn tại m để ....

b/ \(f\left(x\right)\ge0,\forall x\in\left[1;2\right]\)

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\\left[{}\begin{matrix}1< x_1< x_2\\x_1< x_2< 2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-1.f\left(1\right)>0\\\dfrac{x_1+x_2}{2}-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}-1.f\left(2\right)>0\\\dfrac{x_1+x_2}{2}-2< 0\end{matrix}\right.\end{matrix}\right.\)

\(\left(1\right)\left\{{}\begin{matrix}-1+2\sqrt{m}-1-m+\sqrt{m}< 0\\\sqrt{m}-\dfrac{1}{2}-1>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m-3\sqrt{m}+2>0\\\sqrt{m}>\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}0< m< 1\\m>2\end{matrix}\right.\\m>\dfrac{9}{4}\end{matrix}\right.\Leftrightarrow m>\dfrac{9}{4}\)

\(\left(2\right)\left\{{}\begin{matrix}-4+4\sqrt{m}-2-m+\sqrt{m}< 0\\\sqrt{m}-\dfrac{1}{2}-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m-5\sqrt{m}+6>0\\\sqrt{m}< \dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}0< m< 2\\m>3\end{matrix}\right.\\0\le m< \dfrac{25}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< m< 2\\3< m< \dfrac{25}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}m>\dfrac{9}{4}\\0< m< 2\\3< m< \dfrac{25}{4}\end{matrix}\right.\)

\(x^2-\left(m+3\right)x+2m+2=0\Leftrightarrow\left(x-2\right)\left(x-m-1\right)=0\left(1\right)\)

\(\Delta=\left(m+3\right)^2-4\left(2m+2\right)=m^2+6m+9-8m-8=m^2-2m+1=\left(m-1\right)^2\ge0\left(\forall m\right)\)

\(\Rightarrow\left(1\right)\) \(luôn\) \(có\) \(nghiệm\) \(\forall m\)

\(\left(1\right)\Rightarrow\left[{}\begin{matrix}x=2\in(-\text{∞};3]\\x=m+1\end{matrix}\right.\)

\(\left(1\right)\) \(có\) \(đúng\) \(1\) \(nghiệm\) \(\in(-\text{∞};3]\)  \(\Leftrightarrow\left[{}\begin{matrix}x=m+1=2\\x=m+1>3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}m=1\\m>2\end{matrix}\right.\)

\(\Rightarrow\left\{1\right\}\cup\left(2;+\text{∞}\right)\)

\(\)

Phương trình có 2 nghiệm pb khi:

\(\Delta'=\left(m+1\right)^2-m^2>0\Leftrightarrow2m+1>0\)

\(\Rightarrow m>-\dfrac{1}{2}\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x_1+x_2-2}{2}=m\\x_1x_2=m^2\end{matrix}\right.\)

\(\Rightarrow x_1x_2=\left(\dfrac{x_1+x_2-2}{2}\right)^2\)

Đây là hệ thức liên hệ 2 nghiệm ko phụ thuộc m

a,Phương trình có 2 nghiệm pb khi: \(\Delta'>0\Rightarrow\left(m+1\right)^2-m^2>0\Leftrightarrow2m+1>0\Leftrightarrow m>\dfrac{-1}{2}\)