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△=[-2(1-m)]2-4(m2+3)
=4-8m+4m2-4m2-12
=-8-8m
De phuong trinh co 2 nghiem x1,x2 thì :△>=0
=>-8-8m≥0 =>m≤-1
Theo Viet {x1+x2=2-2m ;x1x2=m2+3
=> A=2(2-2m)-m2-3
=4-4m-m2-3
=-m2-4m+1 =-(m2+4m-1)
=-[(m+2)2-5] =-(m+2)2+5
Vì (m+2)2≥0∀m =>-(m+2)2≤0
=>-(m+2)2+5≤5
Vậy GTLN của A là 5 khi m=-2
\(\left\{{}\begin{matrix}m\ne0\\\Delta'>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\ne0\\m< 1\end{matrix}\right.\)
Khi đó \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2m+2}{m}\\x_1x_2=\dfrac{m+3}{m}\end{matrix}\right.\)
\(x_1^3+x_2^3-2\left(x_1+x_2\right)=0\Leftrightarrow\left(x_1+x_2\right)\left(\left(x_1+x_2\right)^2-3x_1x_2\right)-2\left(x_1+x_2\right)=0\)
\(\Leftrightarrow\left(x_1+x_2\right)\left(\left(x_1+x_2\right)^2-3x_1x_2-2\right)=0\)
TH1: \(x_1+x_2=0\Leftrightarrow\dfrac{2\left(m+1\right)}{m}=0\Rightarrow m=-1\)
TH2: \(\left(x_1+x_2\right)^2-3x_1x_2-2=0\Leftrightarrow\left(\dfrac{2m+2}{m}\right)^2-\dfrac{3m+9}{m}-2=0\)
\(\Leftrightarrow m^2+m-4=0\Rightarrow\left[{}\begin{matrix}m=\dfrac{-1-\sqrt{17}}{2}\\m=\dfrac{-1+\sqrt{17}}{2}\left(l\right)\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}m=-1\\m=\dfrac{-1-\sqrt{17}}{2}\end{matrix}\right.\)
Bài 2:
a: \(\text{Δ}=\left(4m+2\right)^2-4\left(4m+3\right)\)
\(=16m^2+16m+4-16m-12=16m^2-8\)
Để phương trình có hai nghiệm thì \(2m^2>=1\)
=>\(\left[{}\begin{matrix}m>=\dfrac{1}{\sqrt{2}}\\m< =-\dfrac{1}{\sqrt{2}}\end{matrix}\right.\)
c: \(A=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)
\(=\left(4m+2\right)^3-3\cdot\left(4m+3\right)\left(4m+2\right)\)
\(=64m^3+96m^2+48m+8-3\left(16m^2+20m+6\right)\)
\(=64m^3+96m^2+48m+8-48m^2-60m-18\)
\(=64m^3+48m^2-12m-10\)
Do \(x_1;x_2;x_3\) là nghiệm của pt nên:
\(\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)=0\)
\(\Leftrightarrow\left(x-x_1\right)\left(x^2-\left(x_2+x_3\right)x+x_2x_3\right)=0\)
\(\Leftrightarrow x^3-\left(x_1+x_2+x_3\right)x^2+\left(x_1x_2+x_2x_3+x_1x_3\right)x-x_1x_2x_3=0\)
Mà \(x^3+ax+b=0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2+x_3=0\\x_1x_2+x_2x_3+x_1x_3=a\\x_1x_2x_3=-b\end{matrix}\right.\) \(\Rightarrow x_1+x_2+x_3=0\)