Tìm m để phương trình sau có nghiệm \(x^2+\frac{1}{x^2}+x+\frac{1}{x}-2m=0\)
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6 tháng 7 2017
Để PT có 2 nghiệm phân biệt thì
\(\Delta'=\left(m-2\right)^2-\left(m^2-2m+4\right)>0\)
\(\Leftrightarrow m< 0\)
Theo vi et ta có:
\(\hept{\begin{cases}x_1+x_2=-2m+4\\x_1.x_2=m^2-2m+4\end{cases}}\)
Theo đề bài thì
\(\frac{2}{x_1^2+x_2^2}-\frac{1}{x_1.x_2}=\frac{15}{m}\)
\(\Leftrightarrow\frac{2}{\left(x_1+x_2\right)^2-2x_1.x_2}-\frac{1}{x_1.x_2}=\frac{15}{m}\)
\(\Leftrightarrow\frac{2}{\left(-2m+4\right)^2-2\left(m^2-2m+4\right)}-\frac{1}{m^2-2m+4}=\frac{15}{m}\)
\(\Leftrightarrow\frac{1}{m^2-6m+4}-\frac{1}{m^2-2m+4}=\frac{15}{m}\)
\(\Leftrightarrow15m^4-120m^3+296m^2-480m+240=0\)
Với m < 0 thì VP > 0
Vậy không tồn tại m để thỏa bài toán.
9 tháng 1
a: \(x^2+\left(2m+1\right)x+m^2-3=0\)
\(\text{Δ}=\left(2m+1\right)^2-4\left(m^2-3\right)\)
\(=4m^2+4m+1-4m^2+12=4m+13\)
Để phương trình có nghiệm kép thì 4m+13=0
=>\(m=-\dfrac{13}{4}\)
Thay m=-13/4 vào phương trình, ta được:
\(x^2+\left(2\cdot\dfrac{-13}{4}+1\right)x+\left(-\dfrac{13}{4}\right)^2-3=0\)
=>\(x^2-\dfrac{11}{2}x+\dfrac{121}{16}=0\)
=>\(\left(x-\dfrac{11}{4}\right)^2=0\)
=>x-11/4=0
=>x=11/4
b: TH1: m=2
Phương trình sẽ trở thành \(\left(2+1\right)x+2-3=0\)
=>3x-1=0
=>3x=1
=>\(x=\dfrac{1}{3}\)
=>Khi m=2 thì phương trình có nghiệm kép là x=1/3
TH2: m<>2
\(\text{Δ}=\left(m+1\right)^2-4\left(m-2\right)\left(m-3\right)\)
\(=m^2+2m+1-4\left(m^2-5m+6\right)\)
\(=m^2+2m+1-4m^2+20m-24\)
\(=-3m^2+22m-23\)
Để phương trình có nghiệm kép thì Δ=0
=>\(-3m^2+22m-23=0\)
=>\(m=\dfrac{11\pm2\sqrt{13}}{3}\)
*Khi \(m=\dfrac{11+2\sqrt{13}}{3}\) thì \(x_1+x_2=\dfrac{-m-1}{m-2}=\dfrac{2-2\sqrt{13}}{3}\)
=>\(x_1=x_2=\dfrac{1-\sqrt{13}}{3}\)
*Khi \(m=\dfrac{11-2\sqrt{13}}{3}\) thì \(x_1+x_2=\dfrac{-m-1}{m-2}=\dfrac{2+2\sqrt{13}}{3}\)
=>\(x_1=x_2=\dfrac{1+\sqrt{13}}{3}\)
c: TH1: m=0
Phương trình sẽ trở thành
\(0x^2-\left(1-2\cdot0\right)x+0=0\)
=>-x=0
=>x=0
=>Nhận
TH2: m<>0
\(\text{Δ}=\left(-1+2m\right)^2-4\cdot m\cdot m\)
\(=4m^2-4m+1-4m^2=-4m+1\)
Để phương trình có nghiệm kép thì -4m+1=0
=>-4m=-1
=>\(m=\dfrac{1}{4}\)
Khi m=1/4 thì \(x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left[-1+2m\right]}{m}=\dfrac{-2m+1}{m}\)
=>\(x_1+x_2=\dfrac{-2\cdot\dfrac{1}{4}+1}{\dfrac{1}{4}}=\dfrac{-\dfrac{1}{2}+1}{\dfrac{1}{4}}=\dfrac{1}{2}:\dfrac{1}{4}=2\)
=>\(x_1=x_2=\dfrac{2}{2}=1\)
KN
7 tháng 2 2021
a) Phương trình \(x^2-2mx-2m-1=0\)có các hệ số a = 1; b = - 2m; c = - 2m - 1
\(\Delta=\left(-2m\right)^2-4\left(-2m-1\right)=4m^2+8m+4=4\left(m+1\right)^2\ge0\forall m\)
Vậy phương trình luôn có 2 nghiệm x1, x2 với mọi m (đpcm)
b) Theo Viète, ta có: \(x_1+x_2=2m;x_1x_2=-2m-1\)
Hệ thức \(\frac{x_1}{x_2}+\frac{x_2}{x_1}=\frac{-5}{2}\Leftrightarrow2\left(x_1^2+x_2^2\right)=-5x_1x_2\)
\(\Leftrightarrow2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]=-5x_1x_2\)hay \(2\left(4m^2+4m+2\right)=10m+5\Leftrightarrow8m^2-2m-1=0\)\(\Leftrightarrow\orbr{\begin{cases}m=\frac{1}{2}\\m=-\frac{1}{4}\end{cases}}\)
Vậy \(m=\frac{1}{2}\)hoặc \(m=-\frac{1}{4}\)thì phương trình có 2 nghiệm x1, x2 thỏa mãn\(\frac{x_1}{x_2}+\frac{x_2}{x_1}=\frac{-5}{2}\)
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