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Bài 3:
a: TH1: m=-2
=>-2(-2-1)x+4<0
=>6x+4<0
=>x<-4/6(loại)
TH2: m<>-2
\(\text{Δ}=\left(2m-2\right)^2-16\left(m+2\right)\)
=4m^2-8m+4-16m-32
=4m^2-24m-28
Để BPT vô nghiệm thì \(\left\{{}\begin{matrix}4m^2-24m-28< =0\\m+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1< =m< =7\\m>-2\end{matrix}\right.\Leftrightarrow-1< =m< =7\)
b: TH1: m=3
=>5x-4>0
=>x>4/5(loại)
TH2: m<>3
Δ=(m+2)^2-4*(-4)(m-3)
\(=m^2+4m+4+16m-48=m^2+20m-44\)
Để bất phương trình vô nghiệm thì
\(\left\{{}\begin{matrix}m^2+20m-44< =0\\m-3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-22< =m< =2\\m< 3\end{matrix}\right.\Leftrightarrow-22< =m< =2\)
a) tử x^2 -8x +20 =(x-4)^2 +4 >0 mọi x => cần
mẫu <0 với mọi x
cần m<0
đủ (m+1)^2 -m(9m+4) <0
<=> m^2 +2m -1 >0
del(m) =1 +1 =2
m <=(-1 -can2)/2
Vô nghiệm với mọi x?
a/ \(\Leftrightarrow\left\{{}\begin{matrix}m-3< 0\\\Delta\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< 3\\\left(m+2\right)^2+16\left(m-3\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow m^2+20m-44\le0\)
\(\Leftrightarrow-22\le m\le2\)
b/ \(\Leftrightarrow\left\{{}\begin{matrix}m< 0\\\Delta'< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< 0\\\left(m-1\right)^2-4m< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< 0\\3-2\sqrt{2}< m< 3+2\sqrt{2}\end{matrix}\right.\)
=> ko tồn tại m thoả mãn
c/ \(\Leftrightarrow\left\{{}\begin{matrix}m^2+2m-3>0\\\Delta'\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>1\\m< -3\end{matrix}\right.\\\left(m-1\right)^2-\left(m^2+2m-3\right)\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>1\\m< -3\end{matrix}\right.\\m\ge1\end{matrix}\right.\Rightarrow m>1\)
a)
\(\left\{{}\begin{matrix}\left(2m-1\right)^2-4\left(m^2-m\right)\ge0\left(1\right)\\\dfrac{1}{m^2-m}>0\left(2\right)\\\dfrac{2m-1}{m^2-m}>0\left(3\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow m^2-m>0\Rightarrow\left[{}\begin{matrix}m< 0\\m>1\end{matrix}\right.\) (I)
Kết hợp \(\left(2\right)\Rightarrow\left(3\right)\Leftrightarrow2m-1>0\Rightarrow m>\dfrac{1}{2}\)(II)
\(\left(1\right)\Leftrightarrow4m^2-4m+1-4m^2+4m=1\ge0\forall m\) (III)
Từ (I) (II) (III) \(\Rightarrow m>1\)
Kết luận nghiệm BPT m>1
b)
\(\left\{{}\begin{matrix}\left(m-2\right)^2-\left(m+3\right)\left(m-1\right)\ge0\left(1\right)\\\dfrac{m-2}{m+3}< 0\left(2\right)\\\dfrac{m-1}{m+3}>0\left(3\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow m^2-4m+4-m^2-2m+3=-6m+7\ge0\Rightarrow m\le\dfrac{7}{6}\)(I)
\(\left(2\right)\Leftrightarrow-3< m< 2\) (2)
\(\left(3\right)\Leftrightarrow\left[{}\begin{matrix}m< -3\\m>1\end{matrix}\right.\)(3)
Nghiệm Hệ BPT là: \(1< m\le\dfrac{7}{6}\)
a)\(\left\{{}\begin{matrix}2m-1>0\Rightarrow m>\dfrac{1}{2}\left(1\right)\\m^2-\left(m-2\right)\left(2m-1\right)< 0\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow m^2-\left(2m^2-m-4m+2\right)=-m^2+5m-2< 0\)
\(m^2-5m+2>0\Rightarrow\left[{}\begin{matrix}m< \dfrac{5-\sqrt{17}}{2}< \dfrac{1}{2}\\m>\dfrac{5+\sqrt{17}}{2}\end{matrix}\right.\)
Nghiệm hệ là
\(m>\dfrac{5+\sqrt{17}}{2}\)
b)\(\left\{{}\begin{matrix}m^2-m-2< 0\left(1\right)\\\left(2m-1\right)^2-4\left(m^2-m-2\right)\le0\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\left(2m-1\right)^2-4\left(m^2-m-2\right)=9< 0,\forall m\).
Suy ra (2) vô nghiệm .
Kết luận hệ vô nghiệm.
a/ \(\Delta'=\left(m-1\right)^2-3\left(m+4\right)< 0\)
\(\Leftrightarrow m^2-5m-11< 0\Leftrightarrow\frac{5-\sqrt{69}}{2}< m< \frac{5+\sqrt{69}}{2}\)
b/ \(\Delta=\left(m+1\right)^2-4\left(2m+7\right)< 0\)
\(\Leftrightarrow m^2-6m-27< 0\Rightarrow-3< m< 9\)
c/ \(\Delta=\left(m-2\right)^2-8\left(-m+4\right)< 0\)
\(\Leftrightarrow m^2+4m-28< 0\Rightarrow-2-4\sqrt{2}< m< -2+4\sqrt{2}\)
d/ \(\left\{{}\begin{matrix}m< 0\\\Delta=\left(m-1\right)^2-4m\left(m-1\right)< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< 0\\\left(m-1\right)\left(-3m-1\right)< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m< 0\\\left[{}\begin{matrix}m< -\frac{1}{3}\\m>1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m< -\frac{1}{3}\)