Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{20a-11}{2012}\\x_1x_2=-1\end{matrix}\right.\)
\(P=\dfrac{3}{2}\left(x_1-x_2\right)^2+2\left(\dfrac{x_1-x_2}{2}-\dfrac{x_1-x_2}{x_1x_2}\right)^2\)
\(=\dfrac{3}{2}\left(x_1-x_2\right)^2+2\left(x_1-x_2\right)^2\left(\dfrac{1}{2}-\dfrac{1}{x_1x_2}\right)^2\)
\(=\dfrac{3}{2}\left(x_1-x_2\right)^2+2\left(x_1-x_2\right)^2\left(\dfrac{1}{2}+1\right)^2\)
\(=6\left(x_1-x_2\right)^2=6\left(x_1+x_2\right)^2-24x_1x_2\)
\(=6\left(\dfrac{20a-11}{2012}\right)^2+24\ge24\)
Dấu "=" xảy ra khi \(a=\dfrac{11}{20}\)
a: \(x^2-x-3m-2=0\)
\(\text{Δ}=\left(-1\right)^2-4\cdot1\cdot\left(-3m-2\right)\)
\(=1+12m+8=12m+9\)
Để phương trình có nghiệm kép thì Δ=0
=>12m+9=0
=>12m=-9
=>\(m=-\dfrac{3}{4}\)
Thay m=-3/4 vào phương trình, ta được:
\(x^2-x-3\cdot\dfrac{-3}{4}-2=0\)
=>\(x^2-x+\dfrac{1}{4}=0\)
=>\(\left(x-\dfrac{1}{2}\right)^2=0\)
=>\(x-\dfrac{1}{2}=0\)
=>\(x=\dfrac{1}{2}\)
b: Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(-1\right)}{1}=1\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{-3m-2}{1}=-3m-2\end{matrix}\right.\)
\(\left(x_1+x_2\right)^2-3x_1x_2\)
\(=1^2-3\left(-3m-2\right)\)
\(=1+9m+6=9m+7\)
c: \(\left(x_1+x_2\right)^2=1^2=1\)
d: \(\left(x_1\right)^2\cdot\left(x_2\right)^2=\left[x_1x_2\right]^2\)
\(=\left(-3m-2\right)^2\)
\(=9m^2+12m+4\)
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 : \(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1.x_2=4m-m^2\end{matrix}\right.\)
\(\Rightarrow\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1.x_2\)
\(\Leftrightarrow\left(x_1-x_2\right)^2=\left(2m+2\right)^2-4.\left(4m-m^2\right)=4m^2+8m+4-16m+4m^2\)
\(\Leftrightarrow\left(x_1-x_2\right)^2=8m^2-8m+4=8\left(m^2+m+\dfrac{1}{4}\right)+2=8\left(m+\dfrac{1}{2}\right)^2+2\ge2\)
\(\Leftrightarrow\left|x_1-x_2\right|\ge\sqrt{2}\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(-2\right)}{4}=\dfrac{1}{2}\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{-1}{4}\end{matrix}\right.\)
\(A=\left(x_1-x_2\right)^2-x_1\left(x_1-\dfrac{1}{2}\right)\)
\(=\left(x_1+x_2\right)^2-4x_1x_2-x_1^2+\dfrac{1}{2}x_1\)
\(=\left(x_1+x_2\right)^2-4x_1x_2-x_1^2+x_1\left(x_1+x_2\right)\)
\(=\left(x_1+x_2\right)^2-4x_1x_2+x_1x_2\)
\(=\left(x_1+x_2\right)^2-3x_1x_2\)
\(=\left(\dfrac{1}{2}\right)^2-3\cdot\dfrac{-1}{4}=\dfrac{1}{4}+\dfrac{3}{4}=1\)
Δ=(2m+2)^2-4(-m-5)
=4m^2+8m+4+4m+20
=4m^2+12m+24
=4(m^2+3m+6)
=4(m^2+2*m*3/2+9/4+15/4)
=4(m+3/2)^2+15>=15
=>PT luôn có 2 nghiệm
(x1-x2)^2-x1(x1+3)-x2(x2+3)=-4
=>(x1+x2)^2-4x1x2-(x1+x2)^2+2x1x2-3(x1+x2)=-4
=>-2(-m-5)-3(2m+2)=-4
=>2m+10-6m-6=-4
=>-4m+4=-4
=>-4m=-8
=>m=2
,có \(ac< 0\)=>pt đã cho luôn có 2 nghiệm phân biệt
vi ét \(=>\left\{{}\begin{matrix}x1+x2=2\\x1x2=-1\end{matrix}\right.\)
a,\(A=\left(x1+x2\right)^2-2x1x2=.....\) thay số tính
b,\(B=\left(x1+x2\right)^3-3x1x2\left(x1+x2\right)=.......\)
c,\(C=x1^{2^2}+x2^{2^2}=\left(x1^2+x2^2\right)^2-2\left(x1x2\right)^2=\left[\left(x1+x2\right)^2-2x1x2\right]^2-2\left(x1x2\right)^2=....\)
\(D=x1x2\left(x1+x2\right)=.....\)
\(x1,x2\ne0=>E=\dfrac{\left(x1+x2\right)^3-3x1x2\left(x1+x2\right)}{x1x2}=...\)
\(F=\sqrt{\left(x1-x2\right)^2}=\sqrt{\left(x1+x2\right)^2-4x1x2}=....\)
\(x1,x2\ne-1=>G=\dfrac{\left(x1+x2\right)^2-2x1x2+x1x2}{x1x2+x1+X2+1}=...\)
\(x1,x2\ne0=>H=\left(\dfrac{x1x2+2}{x2}\right)\left(\dfrac{x1x2+2}{x1}\right)=\dfrac{\left(x1x2+2\right)^2}{x1x2}\)
\(=\dfrac{\left(x1x2\right)^2+4x1x2+4}{x1x2}=..\)