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\(\Delta=\left[-2\left(m+1\right)\right]^2-4\left(m^2-3\right)\)
\(=4m^2+8m+4-4m^2+12=8m+16\)
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)
a: \(\Delta=\left(2m-2\right)^2-4\left(-m-3\right)\)
\(=4m^2-8m+4+4m+12\)
\(=4m^2-4m+16\)
\(=\left(2m-1\right)^2+15>0\)
Do đó: Phương trình luôn có hai nghiệm phân biệt
b: Theo đề, ta có:
\(\left(x_1+x_2\right)^2-2x_1x_2>=10\)
\(\Leftrightarrow\left(2m-2\right)^2-2\left(-m-3\right)>=10\)
\(\Leftrightarrow4m^2-8m+4+2m+6-10>=0\)
\(\Leftrightarrow4m^2-6m>=0\)
=>m<=0 hoặc m>=3/2
△'=(-2)2-1(m-1)
=4-m+1
=5-m
Để PT có 2 no pb thì △'>0
⇒5-m>0
⇒m<5
theo vi-ét ta có
\(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=m-1\end{matrix}\right.\)
mà: \(x^2_1x_2+x_1x_2^2-2\left(x_1+x_2\right)=0\)
⇔\(\left(x_1x_2\right)\left(x_1+x_2\right)-2\left(x_1+x_2\right)=0\)
⇔\(\left(m-1\right)4-2\cdot4=0\)
⇔\(4m-4-8=0\)
⇔4m-12=0
⇔4m=12
⇔m=3
Vậy ...
PT có 2 nghiệm khi:
\(\Delta=\left(m-1\right)^2-4\left(m-1\right)=\left(m-1\right)\left(m-5\right)\ge0\\ \Rightarrow\left[{}\begin{matrix}m< 1\\m>5\end{matrix}\right.\)
Theo Vi-ét: $\begin{cases} x_1+x_2=m-1\\ x_1x_2=m-1 \end{cases}$
Ta có $x_1+2x_2+x_1x_2=m$
\(\Leftrightarrow\left(x_1+ x_2\right)+x_1x_2+x_2=m\\ \Leftrightarrow m-1+x_2+m-1=m\\ \Leftrightarrow x_2=-m+2\)
Mà \(x_1+x_2=m-1\Leftrightarrow x_1=m-1+m-2=2m-3\)
Thay vào $x_1x_2=m-1$
\(\Leftrightarrow\left(2m-3\right)\left(-m+2\right)=m-1\\ \Leftrightarrow2m^2-6m+5=0\left(\text{vô nghiệm}\right)\)
Vậy không có giá trị của \(m\) thỏa mãn
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: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-2\end{matrix}\right.\)
\(\dfrac{\left(x_1^2-2\right)\left(x_2^2-2\right)}{\left(x_1-1\right)\left(x_2-1\right)}=4\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1^2+x_2^2\right)+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)
\(\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1+x_2\right)^2+4x_1x_2+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)
\(\Rightarrow\dfrac{\left(m-2\right)^2-2m^2+4\left(m-2\right)+4}{m-2-m+1}=4\)
\(\Rightarrow-m^2=-4\Rightarrow m=\pm2\)
\(x^2-x+1-m=0\)
Theo Vi - ét, ta có :
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=1\\x_1x_2=\dfrac{c}{a}=1-m\end{matrix}\right.\)
Ta có :
\(5\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}\right)-x_1x_2+4=0\)
\(\Leftrightarrow5\left(\dfrac{x_2+x_1}{x_1x_2}\right)-x_1x_2+4=0\)
\(\Leftrightarrow5\left(\dfrac{1}{1-m}\right)-\left(1-m\right)+4=0\)
\(\Leftrightarrow\dfrac{5}{1-m}-1+m+4=0\)
\(\Leftrightarrow\dfrac{5}{1-m}+m+3=0\)
\(\Leftrightarrow\dfrac{5+m\left(1-m\right)+3\left(1-m\right)}{1-m}=0\)
\(\Leftrightarrow5+m-m^2+3-3m=0\)
\(\Leftrightarrow-m^2-2m+8=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}m=2\\m=-4\end{matrix}\right.\)
a)
Ta có: \(\Delta=\left[-2\left(m+2\right)\right]^2-4\cdot1\cdot\left(m-3\right)\)
\(=\left(-2m-4\right)^2-4\left(m-3\right)\)
\(=4m^2+16m+16\ge0\forall x\)
Suy ra: Phương trình \(x^2-2\left(m+2\right)x+m-3=0\) luôn có nghiệm với mọi m
Áp dụng hệ thức Viet, ta được:
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+2\right)=2m+4\\x_1\cdot x_2=m-3\end{matrix}\right.\)
Ta có: \(\left(2x_1+1\right)\left(2x_2+1\right)=8\)
\(\Leftrightarrow4\cdot x_1x_2+2\cdot\left(x_1+x_2\right)+1=8\)
\(\Leftrightarrow4\left(m-3\right)+2\left(2m+4\right)+1=8\)
\(\Leftrightarrow4m-12+4m+8+1=8\)
\(\Leftrightarrow8m=8+12-8-1\)
\(\Leftrightarrow8m=11\)
hay \(m=\dfrac{11}{8}\)
Tiếp tục với bài của bạn Nguyễn Lê Phước Thịnh
b)
Ta có: \(x_1^2+x_2^2-3x_1x_2=\left(x_1+x_2\right)^2-5x_1x_2\)
\(\Rightarrow P=4m^2+11m+31=4m^2+2\cdot m\cdot\dfrac{11}{2}+\dfrac{121}{4}+\dfrac{3}{4}\) \(=\left(2m+\dfrac{11}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu bằng xảy ra \(\Leftrightarrow2m+\dfrac{11}{2}=0\Leftrightarrow m=-\dfrac{11}{4}\)
Vậy \(P_{Min}=\dfrac{3}{4}\) khi \(m=-\dfrac{11}{4}\)
b) phương trình có 2 nghiệm \(\Leftrightarrow\Delta'\ge0\)
\(\Leftrightarrow\left(m-1\right)^2-\left(m-1\right)\left(m+3\right)\ge0\)
\(\Leftrightarrow m^2-2m+1-m^2-3m+m+3\ge0\)
\(\Leftrightarrow-4m+4\ge0\)
\(\Leftrightarrow m\le1\)
Ta có: \(x_1^2+x_1x_2+x_2^2=1\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=1\)
Theo viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)\\x_1x_2=\dfrac{c}{a}=m+3\end{matrix}\right.\)
\(\Leftrightarrow\left[-2\left(m-1\right)^2\right]-2\left(m+3\right)=1\)
\(\Leftrightarrow4m^2-8m+4-2m-6-1=0\)
\(\Leftrightarrow4m^2-10m-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m_1=\dfrac{5+\sqrt{37}}{4}\left(ktm\right)\\m_2=\dfrac{5-\sqrt{37}}{4}\left(tm\right)\end{matrix}\right.\Rightarrow m=\dfrac{5-\sqrt{37}}{4}\)
1) \(x^2+\left(m+1\right)x+m=2\) ( 1 )
\(pt\left(1\right)\Leftrightarrow x^2+\left(m+1\right)+m-2=0\)
\(\Delta=b^2-4ac\)
\(\Delta=\left(m-1\right)^2+8\ge8\) \(\forall m\in R\)
\(\Rightarrow\) đpcm
2)
Theo định lý Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}\\x_1x_2=\dfrac{c}{a}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=-m-1\\x_1x_2=m-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x_1+x_2\right)^2=\left(-m-1\right)^2\\2x_1x_2=2m-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2_1+2x_1x_2+x^2_2=m^2+2m+1\\2x_1x_2=2m-4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2_1+x^2_2+2m-4=m^2+2m+1\\2x_1x_2=2m-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2_1+x^2_2=\left(m^2+2m+1\right)-\left(2m-4\right)\\2x_2x_1=2m-4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2_1+x^2_2=m^2+5\\x_1x_2=m-2\end{matrix}\right.\)
Theo yêu cầu đề bài \(\dfrac{2x_1-1}{x_2}+\dfrac{2x_2-1}{x_1}=x_1x_2+\dfrac{55}{x_1x_2}\)
\(\Leftrightarrow\dfrac{2\left(x^2_1+x^2_2\right)-\left(x_1+x_2\right)}{x_1x_2}=x_1x_2+\dfrac{55}{x_1x_2}\)
\(\Leftrightarrow\dfrac{2\left(m^2+5\right)-\left(-m-1\right)}{m-2}=m-2+\dfrac{55}{m-2}\)
\(\Leftrightarrow\dfrac{2m^2+m+11}{m-2}=\dfrac{\left(m-2\right)^2+55}{m-2}\)
\(\Leftrightarrow2m^2+m+11=\left(m-2\right)^2+55\) ( điều kiện \(m\ne2\) )
\(\Leftrightarrow m^2+5m-48=0\)
\(\Delta=b^2-4ac\)
\(\Rightarrow\Delta=217\)
\(\Rightarrow m_{1,2}=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-5\pm\sqrt{217}}{2}\)
1)Ta có: \(\Delta_{x,m}=\left(m+1\right)^2-4\left(m-2\right)=\left(m-1\right)^2+8\ge8>0=>dpcm\)
2)Gọi hai nghiệm là a,b cho dẽ viết:
Đặt \(A=\dfrac{2a-1}{b}+\dfrac{2b-1}{a}-ab-\dfrac{55}{ab}\) rút gọn A trước
\(\Leftrightarrow\dfrac{2\left(a^2+b^2\right)-\left(a+b\right)-\left(ab\right)^2-55}{ab}\)
\(A=\dfrac{2\left(a+b\right)^2-\left(a+b\right)-\left[\left(ab\right)^2+4ab+4\right]-51}{ab}\)
\(A=\dfrac{\left(a+b\right)\left[2\left(a+b\right)-1\right]-\left[\left(ab\right)+2\right]^2-51}{ab}\) (1)
Thay Vi_et vào (1) \(\left\{{}\begin{matrix}a+b=-\left(m+1\right)\\ab=m-2\end{matrix}\right.\)
\(A=\dfrac{\left(m+1\right)\left(2m+3\right)-m^2-51}{m-2}=\dfrac{m^2+5m-48}{m-2}\)
\(A=0\Leftrightarrow\left\{{}\begin{matrix}m\ne2\\m^2+5m-48=0\end{matrix}\right.\)
\(\Delta_m=25+4.48=217\) \(\Leftrightarrow\left\{{}\begin{matrix}m\ne2\\\left[{}\begin{matrix}m_2=\dfrac{-5-\sqrt{217}}{2}\ne2\\m_2=\dfrac{-5+\sqrt{217}}{2}\ne2\end{matrix}\right.\end{matrix}\right.\)
Kết luận: \(m_{1,2}=\dfrac{-5\pm\sqrt{217}}{2}\)