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a: \(\text{Δ}=\left[-\left(m+3\right)\right]^2-4\cdot2\cdot m\)
\(=\left(m+3\right)^2-8m\)
\(=m^2-2m+9=\left(m-1\right)^2+8>0\forall m\)
=>Phương trình (1) luôn có hai nghiệm phân biệt
b: Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{m+3}{2}\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{m}{2}\end{matrix}\right.\)
\(A=\left|x_1-x_2\right|=\sqrt{\left(x_1-x_2\right)^2}\)
\(=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(=\sqrt{\dfrac{1}{4}\left(m+3\right)^2-4\cdot\dfrac{m}{2}}\)
\(=\sqrt{\dfrac{1}{4}\left(m^2+6m+9\right)-2m}\)
\(=\sqrt{\dfrac{1}{4}m^2+\dfrac{3}{2}m+\dfrac{9}{4}-2m}\)
\(=\sqrt{\dfrac{1}{4}m^2-\dfrac{1}{2}m+\dfrac{9}{4}}\)
\(=\sqrt{\dfrac{1}{4}\left(m^2-2m+9\right)}\)
\(=\sqrt{\dfrac{1}{4}\left(m^2-2m+1+8\right)}\)
\(=\sqrt{\dfrac{1}{4}\left(m-1\right)^2+2}>=\sqrt{2}\)
Dấu '=' xảy ra khi m-1=0
=>m=1
\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{20a-11}{2012}\\x_1x_2=-1\end{matrix}\right.\)
\(P=\dfrac{3}{2}\left(x_1-x_2\right)^2+2\left(\dfrac{x_1-x_2}{2}-\dfrac{x_1-x_2}{x_1x_2}\right)^2\)
\(=\dfrac{3}{2}\left(x_1-x_2\right)^2+2\left(x_1-x_2\right)^2\left(\dfrac{1}{2}-\dfrac{1}{x_1x_2}\right)^2\)
\(=\dfrac{3}{2}\left(x_1-x_2\right)^2+2\left(x_1-x_2\right)^2\left(\dfrac{1}{2}+1\right)^2\)
\(=6\left(x_1-x_2\right)^2=6\left(x_1+x_2\right)^2-24x_1x_2\)
\(=6\left(\dfrac{20a-11}{2012}\right)^2+24\ge24\)
Dấu "=" xảy ra khi \(a=\dfrac{11}{20}\)
1.
\(a+b+c=0\) nên pt luôn có 2 nghiệm
\(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-1\end{matrix}\right.\)
\(A=\dfrac{2x_1x_2+3}{x_1^2+x_2^2+2x_1x_2+2}=\dfrac{2x_1x_2+3}{\left(x_1+x_2\right)^2+2}=\dfrac{2\left(m-1\right)+3}{m^2+2}=\dfrac{2m+1}{m^2+2}\)
\(A=\dfrac{m^2+2-\left(m^2-2m+1\right)}{m^2+2}=1-\dfrac{\left(m-1\right)^2}{m^2+2}\le1\)
Dấu "=" xảy ra khi \(m=1\)
2.
\(\Delta=m^2-4\left(m-2\right)=\left(m-2\right)^2+4>0;\forall m\) nên pt luôn có 2 nghiệm pb
Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-2\end{matrix}\right.\)
\(\dfrac{\left(x_1^2-2\right)\left(x_2^2-2\right)}{\left(x_1-1\right)\left(x_2-1\right)}=4\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1^2+x_2^2\right)+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)
\(\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1+x_2\right)^2+4x_1x_2+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)
\(\Rightarrow\dfrac{\left(m-2\right)^2-2m^2+4\left(m-2\right)+4}{m-2-m+1}=4\)
\(\Rightarrow-m^2=-4\Rightarrow m=\pm2\)
\(\Delta=\left[-2\left(m+1\right)\right]^2-4\left(m^2-3\right)\)
\(=4m^2+8m+4-4m^2+12=8m+16\)
Để phương trình có hai nghiệm thì 8m+16>=0
hay m>=-2
Áp dụng hệ thức Vi-et, ta được:
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2-3\end{matrix}\right.\)
Theo đề, ta có: \(x_1^2+x_2^2+1=3x_1x_2\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-5x_1x_2+1=0\)
\(\Leftrightarrow\left(2m+2\right)^2-5\left(m^2-3\right)+1=0\)
\(\Leftrightarrow4m^2+8m+4-5m^2+15+1=0\)
\(\Leftrightarrow-m^2+8m+20=0\)
=>(m-10)(m+2)=0
=>m=10 hoặc m=-2
a, \(\Delta'=\left(m+1\right)^2-\left(m^2-3\right)=m^2+2m+1-m^2+3=2m+4\)
Để pt có 2 nghiệm x1 ; x2 khi \(\Delta'\ge0\Leftrightarrow m\ge-2\)
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=m^2-3\end{matrix}\right.\)
Ta có : \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}+\dfrac{1}{x_1x_2}=3\Leftrightarrow\dfrac{\left(x_1+x_2\right)^2-2x_1x_2+1}{x_1x_2}=3\)
\(\Leftrightarrow\dfrac{4\left(m^2+2m+1\right)-2\left(m^2-3\right)+1}{m^2-3}=3\)
\(\Rightarrow2m^2+8m+11=3m^2-9\Leftrightarrow m^2-8m-20=0\Leftrightarrow m=10;m=-2\)(tm)
Theo vi-et thì ta có:
\(\hept{\begin{cases}x_1+x_2=\frac{3a-1}{2}\\x_1x_2=-1\end{cases}}\)
Từ đây ta có:
\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2=\left(\frac{3a-1}{2}\right)^2-4.1=\left(\frac{3a-1}{2}\right)^2-4\)
Theo đề bài thì
\(P=\frac{3}{2}.\left(x_1-x_2\right)^2+2\left(\frac{x_1-x_2}{2}+\frac{1}{x_1}-\frac{1}{x_2}\right)^2\)
\(=\frac{3}{2}.\left(x_1-x_2\right)^2+2.\left(x_1-x_2\right)^2\left(\frac{1}{2}-\frac{1}{x_1x_2}\right)^2\)
\(=\left(x_1-x_2\right)^2\left(\frac{3}{2}+2.\left(\frac{1}{2}-\frac{1}{x_1x_2}\right)^2\right)\)
\(=\left(\left(\frac{3a-1}{2}\right)^2-4\right)\left(\frac{3}{2}+2.\left(\frac{1}{2}+1\right)^2\right)\)
\(=6\left(\left(\frac{3a-1}{2}\right)^2-4\right)\ge6.4=24\)
Dấu = xảy ra khi \(a=\frac{1}{3}\)
\(x^2-2\left(m+1\right)x+3\left(m+1\right)-3=0\)
\(x^2-2nx+3n+3=\left(x-n\right)^2-\left(n^2-3n+3\right)=0\)\(\left(x-n\right)^2=\left(n-\frac{3}{2}\right)^2+\frac{3}{4}=\frac{\left(2n-3\right)^2+3}{4}>0\forall n\) vậy luôn tồn tại hai nghiệm
\(\orbr{\begin{cases}x_1=\frac{n-\sqrt{\left(2n-3\right)^2+3}}{2}\\x_2=\frac{n+\sqrt{\left(2n-3\right)^2+3}}{2}\end{cases}}\)
a) \(\frac{x_1}{x_2}=\frac{4x_1-x_2}{x_1}\Leftrightarrow\frac{x_1^2-4x_1x_2+x_2^2}{x_1x_2}=0\)
\(x_1x_2=n^2-\frac{\left(2n-3\right)^2+3}{4}=\frac{4n^2-4n^2+12n-9-3}{4}=3n-3\)
với n=1 hay m=0 : Biểu thức cần C/m không tồn tại => xem lại đề