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a)
\(\left\{{}\begin{matrix}x^2+x+5< 0\\x^2-6x+1>0\end{matrix}\right.\)
\(\)Ta có
\(x^2+x+5=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{19}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}>0\)
=> Bất phương trình đàu tiên sai, hệ bất phương trình sai
b)
\(\left\{{}\begin{matrix}2x^2+x-6>0\\3x^2-10x+3\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3\right)\left(x+2\right)>0\\\left(x-3\right)\left(3x-1\right)\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>2\\x< -3\end{matrix}\right.\\\left[{}\begin{matrix}x\le-\dfrac{1}{3}\\x\ge3\end{matrix}\right.\end{matrix}\right.\)
a, hệ\(\Leftrightarrow\)$\left \{ {{x>\frac{1}{2} } \atop {x<m+2}} \right.$
để hệ có nghiệm ⇒ m+2< $\frac{1}{2}$ ⇒ m<$\frac{-3}{2}$
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) \(\left\{\begin{matrix} 2x^2+x-6>0\\ 3x^2-10x+3\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (2x-3)(x+2)>0\\ (x-3)(3x-1)\geq 0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} \left[\begin{matrix} x>\frac{3}{2}\\ x< -2\end{matrix}\right.\\ \left[\begin{matrix} x\geq 3\\ x\leq \frac{1}{3}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow \left[\begin{matrix} x\geq 3\\ x< -2\end{matrix}\right.\)
Xét \(-x^2+2x+3\le0\Leftrightarrow\left[{}\begin{matrix}x\le-1\\x\ge3\end{matrix}\right.\)
Xét \(x+2m-1>0\Leftrightarrow x>-2m+1\)
Hệ đã cho có nghiệm với mọi m (đều chứa khoảng dương vô cùng)
\(\left\{{}\begin{matrix}-x^2+2x+3\le0\\x+2m-1>0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}-1\le x\le3\\x>-2m+1\end{matrix}\right.\)
để pt ....thì \(-2m+1< 3\)
<=>\(-2m< 2\)
<=> \(m>1\)
vậy pt .....
Phải là dấu ngoặc nhọn chứ=0
\(\left\{{}\begin{matrix}2x+3y-6< 0\\x\ge0\\2x-3y-1\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3y-6\le2x+3y-6< 0\\x\ge0\\-3y-1\le2x-3y-1\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3y-6< 0\\-3y-1\le0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y< 2\\y\ge-\dfrac{1}{3}\\x\ge0\end{matrix}\right.\)
=> Miền nghiệm là \([0;2)\)
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:
\(\left\{{}\begin{matrix}\left(2m-1\right)\left(4-4m\right)>0\\\dfrac{-2m+3}{2m-1}=\dfrac{3}{4-4m}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}< m< 1\\\left[{}\begin{matrix}m=\dfrac{3}{4}\\m=\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow m=\dfrac{3}{4}\)
\(\left\{{}\begin{matrix}x^2+4x+3\ge0\\2x^2-x-10\le0\\2x^2-5x+3>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(x+3\right)\ge0\\\left(x+2\right)\left(2x-5\right)\le0\\\left(2x-3\right)\left(x-1\right)>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le-3\end{matrix}\right.\\-2\le x\le\dfrac{5}{2}\\\left[{}\begin{matrix}x< 1\\x>\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}-1\le x\le1\\\dfrac{3}{2}\le x\le\dfrac{5}{2}\end{matrix}\right.\)
là vầy bạn giải ra xong giao nghiệm lại.