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

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(x-4\right)< 0\\\left(m-1\right)x-2>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-1< x< 4\\\left(m-1\right)x-2>0\end{matrix}\right.\left(1\right)\)

TH1: \(m< 1\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}-1< x< 4\\x< \dfrac{2}{m-1}\end{matrix}\right.\)

Yêu cầu bài toán thỏa mãn khi \(\dfrac{2}{m-1}>-1\Leftrightarrow2< -m+1\Leftrightarrow m< -1\)

\(\Rightarrow m< -1\)

TH2: \(m=1\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}-1< x< 4\\-2>0\end{matrix}\right.\left(vn\right)\)

TH3: \(m>1\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}-1< x< 4\\x>\dfrac{2}{m-1}\end{matrix}\right.\)

\(\dfrac{2}{m-1}< 4\Leftrightarrow4m-4>2\Leftrightarrow m>\dfrac{3}{2}\)

\(\Rightarrow m>\dfrac{3}{2}\)

Vậy \(m< -1;m>\dfrac{3}{2}\)

\(\left\{{}\begin{matrix}x^2-3x+2\le0\\mx+1-m\le0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}1\le x\le2\\x\le\dfrac{-1+m}{m}\end{matrix}\right.\)

để hpt trên có nghiệm thì \(\dfrac{-1+m}{m}\le2\) ĐK m ≠ 0

\(< =>m\ge-1\)

Vậy .....

\(x^2-3x+2\le0\Leftrightarrow1\le x\le2\) \(\Rightarrow D_1=\left[1;2\right]\)

Xét \(mx\le m-1\)

- Với \(m=0\) BPT vô nghiệm

- Với \(m>0\Leftrightarrow x\le\dfrac{m-1}{m}\) \(\Rightarrow D_2=(-\infty;\dfrac{m-1}{m}]\)

Hệ có nghiệm khi \(D_1\cap D_2\ne\varnothing\)

\(\Leftrightarrow\dfrac{m-1}{m}\ge1\) \(\Rightarrow\) không tồn tại m thỏa mãn

- Với \(m< 0\Leftrightarrow x\ge\dfrac{m-1}{m}\Rightarrow D_2=[\dfrac{m-1}{m};+\infty)\)

\(D_1\cap D_2\ne\varnothing\Leftrightarrow\dfrac{m-1}{m}\le2\)

\(\Leftrightarrow m-1\ge2m\Rightarrow m\le-1\)

Vậy \(m\le-1\)

a.

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\le m\end{matrix}\right.\)

Hệ có nghiệm duy nhất \(\Leftrightarrow m=2\)

b.

\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2+1\right)x\ge6\\2x\le6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{6}{m^2+1}\\x\le3\end{matrix}\right.\)

Hệ có nghiệm duy nhất \(\Leftrightarrow\dfrac{6}{m^2+1}=3\)

\(\Leftrightarrow m=\pm1\)

c.

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-6x+9\ge x^2+7x+1\\5x\ge2m-8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{8}{13}\\x\ge\dfrac{2m-8}{5}\end{matrix}\right.\)

Pt có nghiệm duy nhất khi \(\dfrac{2m-8}{5}=\dfrac{8}{13}\Leftrightarrow m=\dfrac{72}{13}\)

d.

Hệ có nghiệm duy nhất khi:

TH1:

 \(\left\{{}\begin{matrix}m>0\\\dfrac{m-3}{m}=\dfrac{m-9}{m+3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>0\\m^2-9=m^2-9m\end{matrix}\right.\) \(\Leftrightarrow m=1\)

TH2:

\(\left\{{}\begin{matrix}m+3< 0\\\dfrac{m-3}{m}=\dfrac{m-9}{m+3}\end{matrix}\right.\)

\(\Leftrightarrow m=1\) (ktm)

Vậy \(m=1\)

e.

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2m-1\right)x\ge-2m+3\\\left(4-4m\right)x\le3\end{matrix}\right.\)

Hệ có nghiệm duy nhất khi:

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

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}< m< 1\\\left[{}\begin{matrix}m=\dfrac{3}{4}\\m=\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow m=\dfrac{3}{4}\)

a, hệ\(\Leftrightarrow\)$\left \{ {{x>\frac{1}{2} } \atop {x<m+2}} \right.$

để hệ có nghiệm ⇒ m+2< $\frac{1}{2}$ ⇒ m<$\frac{-3}{2}$