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a, hệ\(\Leftrightarrow\)$\left \{ {{x>\frac{1}{2} } \atop {x<m+2}} \right.$
để hệ có nghiệm ⇒ m+2< $\frac{1}{2}$ ⇒ m<$\frac{-3}{2}$
![](https://rs.olm.vn/images/avt/0.png?1311)
c)
\(\left\{\begin{matrix} -x^2+4x-7< 0\\ x^2-2x-1\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x^2-4x+7>0\\ x^2-2x+1\geq 2\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} (x-2)^2+3>0\\ (x-1)^2-2\geq 0\end{matrix}\right.\Leftrightarrow (x-1)^2-2\geq 0\Leftrightarrow \left[\begin{matrix} x-1\geq \sqrt{2}\\ x-1\leq -\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x\geq \sqrt{2}+1\\ x\leq 1-\sqrt{2}\end{matrix}\right.\)
d)
\(\left\{\begin{matrix} -2x^2-5x+4< 0\\ -x^2-3x+10>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 2x^2+5x-4>0\\ (2-x)(x+5)>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 2(x+\frac{5}{4})^2-\frac{57}{8}>0\\ (2-x)(x+5)>0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} (x+\frac{5}{4}-\frac{\sqrt{57}}{4})(x+\frac{5}{4}+\frac{\sqrt{57}}{4})>0\\ (2-x)(x+5)>0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} \left[\begin{matrix} x>\frac{-5+\sqrt{57}}{4}\\ x< \frac{-5-\sqrt{57}}{4}\end{matrix}\right.\\ -5< x< 2\end{matrix}\right.\) \(\Rightarrow \left[\begin{matrix} -5< x< \frac{-5-\sqrt{57}}{4}\\ \frac{\sqrt{57}-5}{4}< x< 2\end{matrix}\right.\)
a)
\(\left\{\begin{matrix} 2x^2+9x+7>0\\ x^2+x-6< 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (x+1)(2x+7)>0\\ (x-2)(x+3)< 0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} \left[\begin{matrix} x>-1\\ x< \frac{-7}{2}\end{matrix}\right.\\ -3< x< 2\end{matrix}\right.\Rightarrow -1< x< 2\)
b) \(\left\{\begin{matrix} 2x^2+x-6>0\\ 3x^2-10x+3\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (2x-3)(x+2)>0\\ (x-3)(3x-1)\geq 0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} \left[\begin{matrix} x>\frac{3}{2}\\ x< -2\end{matrix}\right.\\ \left[\begin{matrix} x\geq 3\\ x\leq \frac{1}{3}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow \left[\begin{matrix} x\geq 3\\ x< -2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(\left\{{}\begin{matrix}\left(2m-1\right)\left(4-4m\right)>0\\\dfrac{-2m+3}{2m-1}=\dfrac{3}{4-4m}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}< m< 1\\\left[{}\begin{matrix}m=\dfrac{3}{4}\\m=\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow m=\dfrac{3}{4}\)
Lời giải:
HBPT tương đương với \(\left\{\begin{matrix} (4x-1)(x+1)>0\\ x\leq 3\end{matrix}\right.\) \(\Leftrightarrow \left\{\begin{matrix} x>\frac{1}{4} \text{hoặc} x< -1\\ x\leq 3\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} \frac{1}{4}< x\leq 3\\ x< -1\end{matrix}\right.\)