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\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{20a-11}{2012}\\x_1x_2=-1\end{matrix}\right.\)
\(P=\dfrac{3}{2}\left(x_1-x_2\right)^2+2\left(\dfrac{x_1-x_2}{2}-\dfrac{x_1-x_2}{x_1x_2}\right)^2\)
\(=\dfrac{3}{2}\left(x_1-x_2\right)^2+2\left(x_1-x_2\right)^2\left(\dfrac{1}{2}-\dfrac{1}{x_1x_2}\right)^2\)
\(=\dfrac{3}{2}\left(x_1-x_2\right)^2+2\left(x_1-x_2\right)^2\left(\dfrac{1}{2}+1\right)^2\)
\(=6\left(x_1-x_2\right)^2=6\left(x_1+x_2\right)^2-24x_1x_2\)
\(=6\left(\dfrac{20a-11}{2012}\right)^2+24\ge24\)
Dấu "=" xảy ra khi \(a=\dfrac{11}{20}\)
\(\Delta'=\left(2m+1\right)^2-\left(4m^2+4m\right)=1>0;\forall m\Rightarrow\) pt luôn có 2 nghiệm pb
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(2m+1\right)\\x_1x_2=4m^2+4m\end{matrix}\right.\)
\(\left|x_1-x_2\right|=x_1+x_2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2\ge0\\\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(2m+1\right)\ge0\\-2x_1x_2=2x_1x_2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ge-\dfrac{1}{2}\\x_1x_2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ge-\dfrac{1}{2}\\4m^2+4m=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}m=0\\mm=-1< -\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)
Lời giải:
$\Delta'=(m+1)^2-(4m-m^2)=2m^2-2m+1=2(m-0,5)^2+0,5>0$ với mọi $m$ nên pt luôn có 2 nghiệm pb với mọi $m$
Áp dụng định lý Viet: \(\left\{\begin{matrix} x_1+x_2=2(m+1)\\ x_1x_2=4m-m^2\end{matrix}\right.\)
Khi đó:
\(P=|x_1-x_2|=\sqrt{(x_1-x_2)^2}=\sqrt{(x_1+x_2)^2-4x_1x_2}\)
\(=\sqrt{4(m+1)^2-4(4m-m^2)}=\sqrt{4(2m^2-2m+1)}\)
\(=2\sqrt{2(m-0,5)^2+0,5}\geq 2\sqrt{0,5}\)
Vậy $P_{\min}=2\sqrt{0,5}=\sqrt{2}$. Giá trị này đạt tại $m=0,5$
Theo Vi-et : \(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1.x_2=4m-m^2\end{matrix}\right.\)
\(\Rightarrow\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1.x_2\)
\(\Leftrightarrow\left(x_1-x_2\right)^2=\left(2m+2\right)^2-4.\left(4m-m^2\right)=4m^2+8m+4-16m+4m^2\)
\(\Leftrightarrow\left(x_1-x_2\right)^2=8m^2-8m+4=8\left(m^2+m+\dfrac{1}{4}\right)+2=8\left(m+\dfrac{1}{2}\right)^2+2\ge2\)
\(\Leftrightarrow\left|x_1-x_2\right|\ge\sqrt{2}\)
a)Có ac=-1<0
=>pt luôn có hai nghiệm trái dấu
b)Do x1;x2 là hai nghiệm của pt
=> \(\left\{{}\begin{matrix}x_1^2-mx_1-1=0\\x_2^2-mx_2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_1^2-1=mx_1\\x_2^2-1=mx_2\end{matrix}\right.\)
=>\(P=\dfrac{mx_1+x_1}{x_1}-\dfrac{mx_2+x_2}{x_2}\)\(=m+1-\left(m+1\right)=0\)
Ta có để phương trình có nghiệm thì:
\(\Delta=k^2-4\ge0\)
\(\Leftrightarrow k\ge2;k\le-2\)
Theo đề thì ta có
\(\left(\frac{x_1}{x_2}\right)^2+\left(\frac{x_2}{x_1}\right)^2\ge3\)
\(\Leftrightarrow x_1^4+x_2^4-3\left(x_1x_2\right)^2\ge0\)
\(\Leftrightarrow\left(\left(x_1+x_2\right)^2-2x_1x_2\right)^2-5x_1x_2\ge0\)
\(\Leftrightarrow\left(4k^2-4\right)^2-5.4^2\ge0\)
Làm nốt
\(\left|k\right|\ge2\)
\(P=\left(\frac{x_1}{x_2}\right)^2+\left(\frac{x_2}{x_1}\right)^2=\left(\frac{x_1}{x_2}+\frac{x_2}{x_1}\right)^2-2=\left(\frac{\left(x_1+x_2\right)^2}{x_1x_2}-2\right)^2-2\\ \)
\(P=\left(\frac{\left(2k\right)^2}{4}-2\right)^2-2=\left(k^2-2\right)^2-2\)
\(P\ge3\Rightarrow\left(k^2-2\right)^2\ge5\Leftrightarrow\orbr{\begin{cases}k^2-2\le-\sqrt{5}\left(l\right)\\k^2-2\ge\sqrt{5}\left(n\right)\end{cases}}\)
\(\orbr{\begin{cases}k\le-\sqrt{2+\sqrt{5}}\\k\ge\sqrt{2+\sqrt{5}}\end{cases}}\)
\(\Delta'=\left(m-3\right)^2-\left(-6m-7\right)=m^2+16>0\)
Vậy pt có 2 nghiệm pb
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-3\right)\\x_1x_2=-6m-7\end{matrix}\right.\)
\(C=4\left(m-3\right)^2+8\left(-6m-7\right)\)
\(=4m^2-24m+36-48m-56=4m^2-72m-20\)
\(=4\left(m^2-18m+81-81\right)-20=4\left(m-9\right)^2-344\ge-344\)
Dấu ''='' xảy ra khi m = 9
\(\Delta'=\left[-\left(m+1\right)\right]^2-\left(m^2+m\right)=m^2+2m+1-m^2-m\)
\(=m+1\)
pt có nghiệm x1,x2 \(< =>m+1\ge0< =>m\ge-1\)
vi ét \(=>\left\{{}\begin{matrix}x1+x2=2m+2\\x1x2=m^2+m\end{matrix}\right.\)
a,\(=>2m+2=m^2+m< =>m^2-m-2=0\)
\(a-b+c=0=>\left[{}\begin{matrix}m1=-1\\m2=2\end{matrix}\right.\left(tm\right)\)
b,\(< =>3\left(2m+2\right)-2\left(m^2+m\right)-1=0\)
\(< =>-2m^2+4m+5=0\)
\(ac< 0\) pt có 2 nghiệm pbiet \(=>\left[{}\begin{matrix}m1=...\\m2=...\end{matrix}\right.\) thay số vào tính m1,m2 đối chiếu đk
