Cho phương trình
\(2x^2+\left(m-1\right)x-2\)=0
Tim m để
\(\left(x_1+\frac{1}{2}x^2_1-x^3_1\right)\left(x_2+\frac{1}{2}x^2_2-x^3_2\right)=4\)
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\(x^2-4x-6=0\)
\(\text{Δ}=\left(-4\right)^2-4\cdot1\cdot\left(-6\right)=16+24=40>0\)
=>Phương trình này có hai nghiệm phân biệt
Theo vi-et, ta có:
\(x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left(-4\right)}{1}=4;x_1\cdot x_2=\dfrac{c}{a}=\dfrac{-6}{1}=-6\)
\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)
\(=4^2-2\cdot\left(-6\right)=16+12=28\)
\(B=\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{x_1\cdot x_2}=\dfrac{4}{-6}=-\dfrac{2}{3}\)
\(C=x_1^3+x_2^3\)
\(=\left(x_1+x_2\right)^3-3\cdot x_1\cdot x_2\cdot\left(x_1+x_2\right)\)
\(=4^3-3\cdot4\cdot\left(-6\right)=64+72=136\)
\(D=\left|x_1-x_2\right|\)
\(=\sqrt{\left(x_1-x_2\right)^2}\)
\(=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(=\sqrt{4^2-4\cdot\left(-6\right)}=\sqrt{16+24}=\sqrt{40}=2\sqrt{10}\)
b, Ta có : \(0\le x\le1\)
\(\Rightarrow-2\le x-2\le-1< 0\)
Ta có : \(y=f\left(x\right)=2\left(m-1\right)x+\dfrac{m\left(x-2\right)}{\left(2-x\right)}\)
\(=2\left(m-1\right)x-m< 0\)
TH1 : \(m=1\) \(\Leftrightarrow m>0\)
TH2 : \(m\ne1\) \(\Leftrightarrow x< \dfrac{m}{2\left(m-1\right)}\)
Mà \(0\le x\le1\)
\(\Rightarrow\dfrac{m}{2\left(m-1\right)}>1\)
\(\Leftrightarrow\dfrac{m-2\left(m-1\right)}{2\left(m-1\right)}>0\)
\(\Leftrightarrow\dfrac{2-m}{m-1}>0\)
\(\Leftrightarrow1< m< 2\)
Kết hợp TH1 => m > 0
Vậy ...
\(x^2-2\left(m-1\right)x-m^3+\left(m+1\right)^2=0\)
Để pt có hai nghiệm thỏa mãn
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta\ge0\\x_1+x_2=2\left(m-1\right)\le4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m\left(m-2\right)\left(m+2\right)\ge0\\m\le3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}m\in\left[-2;0\right]\cup\left(2;+\infty\right)\cup\left\{2\right\}\\m\le3\end{matrix}\right.\)\(\Rightarrow m\in\left[-2;0\right]\cup\left[2;3\right]\)
\(P=x^3_1+x_2^3+x_1x_2\left(3x_1+3x_2+8\right)\)
\(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)+3x_1x_1\left(x_1+x_2\right)+8x_1x_2\)
\(=8\left(m-1\right)^3+8\left(-m^3+m^2+2m+1\right)\)
\(=-16m^2+40m\)
Vẽ BBT với \(f\left(m\right)=-16m^2+40m\) ;\(m\in\left[-2;0\right]\cup\left[2;3\right]\)
Tìm được \(f\left(m\right)_{min}=-144\Leftrightarrow m=-2\)
\(f\left(m\right)_{max}=16\Leftrightarrow m=2\)
\(\Rightarrow P_{max}=16;P_{min}=-144\)
Vậy....
Ta có: \(\frac{c}{a}=-\frac{2}{2}=-1< 0\)
=> Phương trình luôn có 2 ngiệm trái dấu \(x_1;x_2\)
Theo định lí viet: \(x_1x_2=-1;x_1+x_2=\frac{1-m}{2}\)
Ta có: \(\left(x_1+\frac{1}{2}x^2_1-x^3_1\right)\left(x_2+\frac{1}{2}x^2_2-x^3_2\right)=4\)
<=> \(x_1x_2\left(x_1^2-\frac{1}{2}x_1-1\right)\left(x_2^2-\frac{1}{2}x_2-x_2\right)=4\)
<=> \(\left(2x_1^2-x_1-2\right)\left(2x_2^2-x_2-2\right)=-16\)
<=> \(\left(2x_1x_2\right)^2-2x_1^2x_2-4x_1^2-2x_1x_2^2+x_1x_2+2x_2-4x_2^2+2x_2+4=-16\)
<=> \(4+2x_1-4x_1^2+2x_2-1+2x_2-4x_2^2+2x_2+4=-16\)
<=> \(4x_1^2+4x_2^2-4x_1-4x_2=23\)
<=> \(4\left(x_1+x_2\right)^2-4\left(x_1+x_2\right)=15\)
<=> \(\orbr{\begin{cases}x_1+x_2=\frac{5}{2}\\x_1+x_2=-\frac{3}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1-m}{2}=\frac{5}{2}\\\frac{1-m}{2}=-\frac{3}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}m=-4\\m=4\end{cases}}\)
Vậy:....