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\(\Delta=\left(3m+2\right)^2-12m=9m^2+4>0\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-3m-2\\x_1x_2=3m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1+1+x_2+1=-3m\\x_1x_2+x_1+x_2+1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1+1+x_2+1=-3m\\\left(x_1+1\right)\left(x_2+1\right)=-1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x_1+1=a\\x_2+1=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=-3m\\ab=-1\end{matrix}\right.\)
\(Q=a^4+b^4\ge2a^2b^2=2\)
Dấu "=" xảy ra khi \(a^2=b^2\Rightarrow\left[{}\begin{matrix}a=b\left(loại\right)\\a=-b\end{matrix}\right.\)
\(\Rightarrow-3m=0\Rightarrow m=0\)
\(\Delta=\left(2m-2\right)^2-4\cdot2\cdot\left(m^2-1\right)\)
\(=4m^2-8m+4-8m^2+8\)
\(=-4m^2-8m+12\)
Để phương trình có hai nghiệm phân biệt thì -4m^2-8m+12>0
=>4m^2+8m-12<0
=>m^2+2m-3<0
=>(m+3)(m-1)<0
=>-3<m<1
\(A=\left(x_1+x_2\right)^2-4x_1x_2\)
\(=\left(\dfrac{2m-2}{2}\right)^2-4\cdot\dfrac{m^2-1}{2}\)
\(=\left(m-1\right)^2-2\left(m^2-1\right)\)
\(=m^2-2m+1-2m^2+2=-m^2-2m+3\)
\(=-\left(m^2+2m-3\right)\)
\(=-\left(m^2+2m+1-4\right)\)
\(=-\left(m+1\right)^2+4< =4\)
Dấu = xảy ra khi m=-1
1: \(\text{Δ}=\left(-m\right)^2-4\left(m-2\right)=m^2-4m+8=\left(m-2\right)^2+4>0\)
=>Phương trình luôn có hai nghiệm phân biệt
Theo đề, ta có: m-2<0
=>m<2
2: \(\Leftrightarrow\dfrac{x_1^2+1}{x_1}\cdot\dfrac{x_2^2+1}{x_2}=9\)
\(\Leftrightarrow\dfrac{\left(x_1\cdot x_2\right)^2+\left(x_1+x_2\right)^2-2x_1x_2+1}{x_1x_2}=9\)
\(\Leftrightarrow\dfrac{\left(m-2\right)^2+\left(-m\right)^2-2\left(m-2\right)+1}{m-2}=9\)
\(\Leftrightarrow m^2-4m+4+m^2-2m+4+1=9m-18\)
\(\Leftrightarrow2m^2-6m+9-9m+18=0\)
=>2m^2-15m+27=0
hay \(m\in\varnothing\)
3: =>m=0
\(\Delta=\left(2m+1\right)^2-4\left(m^2+m-2\right)=9>0\)
\(\Rightarrow\) phương trình luôn có 2 nghiệm phân biệt: \(\left\{{}\begin{matrix}x_1=\frac{2m+1+3}{2}=m+2\\x_2=\frac{2m+1-3}{2}=m-1\end{matrix}\right.\)
Để phương trình có 2 nghiệm âm phân biệt:
\(\Rightarrow x_1< 0\Rightarrow m+2< 0\Rightarrow m< -2\)
Khi đó:
\(A=x_1\left(x_2+5\right)=\left(m+2\right)\left(m-1+5\right)=\left(m+2\right)\left(m+4\right)\)
\(A=m^2+6m+8=\left(m+3\right)^2-1\ge-1\)
\(\Rightarrow A_{min}=-1\) khi \(m+3=0\Leftrightarrow m=-3< -2\) (thỏa mãn)
Để pt có 2 nghiệm trái dấu \(\Leftrightarrow ac< 0\Leftrightarrow-\left(m^2-4\right)< 0\Rightarrow\left[{}\begin{matrix}m>2\\m< -2\end{matrix}\right.\)
Do \(x_1< x_2\Rightarrow x_1< 0< x_2\Rightarrow\left\{{}\begin{matrix}\left|x_1\right|=-x_1\\\left|x_2\right|=x_2\end{matrix}\right.\)
\(\left|x_1\right|>\left|x_2\right|\Leftrightarrow-x_1>x_2\Leftrightarrow x_1+x_2< 0\Leftrightarrow\frac{-\left(m+3\right)}{-1}< 0\Rightarrow m< -3\)
a) △ = \(m^2-28\ge0\)\(\Leftrightarrow\left[{}\begin{matrix}m\ge\sqrt{28}\\m\le-\sqrt{28}\end{matrix}\right.\)
Theo Vi-ét \(\left\{{}\begin{matrix}x_1+x_2=-m\\x_1x_2=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_1^2+x_2^2+2x_1x_2=m^2\\x_1x_2=7\end{matrix}\right.\)
\(\Rightarrow m^2=24\)\(\Leftrightarrow\left[{}\begin{matrix}m=\sqrt{24}\\m=-\sqrt{24}\end{matrix}\right.\)(không thỏa mãn)
b) △ = \(4-4\left(m+2\right)\ge0\)\(\Leftrightarrow m\le-1\)
Theo Vi-ét \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m+2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_1^2+x_2^2+2x_1x_2=4\\x_1x_2=m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x_2-x_1\right)^2+4x_1x_2=4\\x_1x_2=m+2\end{matrix}\right.\)
\(\Rightarrow4+4\left(m+2\right)=4\)\(\Leftrightarrow m=-2\)(thỏa mãn)
c) △ = \(\left(m-1\right)^2-4\left(m+6\right)\)\(\ge0\)\(\Leftrightarrow m^2-2m+1-4m-24\ge0\)
\(\Leftrightarrow m^2-6m-23\ge0\)
\(\Leftrightarrow\left(m-3\right)^2\ge32\)\(\Leftrightarrow\left[{}\begin{matrix}m\ge\sqrt{32}+3\\m\le-\sqrt{32}+3\end{matrix}\right.\)
Theo Vi-ét \(\left\{{}\begin{matrix}x_1+x_2=1-m\\x_1x_2=m+6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_1^2+x_2^2+2x_1x_2=m^2-2m+1\\x_1x_2=m+6\end{matrix}\right.\)
\(\Rightarrow10+2\left(m+6\right)=m^2-2m+1\)
\(\Leftrightarrow m^2-4m-21=0\)\(\Leftrightarrow\left(m+3\right)\left(m-7\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}m=7\\m=-3\end{matrix}\right.\)\(\Leftrightarrow m=-3\)(thỏa mãn)
mấy câu kia cũng dùng Vi-ét xử tiếp nha