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\(\Delta'=b'^2-ac=\left[-\left(m-2\right)\right]^2-1.\left(m^2+2m-3\right)=-6m+7\)
Để pt có 2 no thì \(\Delta'>0\Leftrightarrow-6m+7>0\Leftrightarrow m< \frac{7}{6}\)
Theo Vi-ét ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-2\right)\\x_1.x_2=m^2+2m-3\end{matrix}\right.\)
Mặt khác: \(\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{5}\Leftrightarrow5\left(x_1+x_2\right)=x_1.x_2\left(x_1+x_2\right)\Leftrightarrow\left(x_1+x_2\right)\left(5-x_1.x_2\right)=0\)
Do đó: \(2\left(m-2\right)\left(5-m^2-2m+3\right)=0\Leftrightarrow\left[{}\begin{matrix}m=2\left(loại\right)\\m=-4\end{matrix}\right.\)
Vậy khi m=-4 thì thỏa mãn...
Xét \(\Delta=4\left(m-1\right)^2-4.\left(-3\right)=4\left(m-1\right)^2+12>0\forall m\)
=>Pt luôn có hai nghiệm pb
Theo viet:\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1.x_2=-3\ne0\forall m\end{matrix}\right.\)
Có \(\dfrac{x_1}{x_2^2}+\dfrac{x_2}{x_1^2}=m-1\)
\(\Leftrightarrow x_1^3+x_2^3=\left(m-1\right)x_1^2.x_2^2\)
\(\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=\left(m-1\right).\left(-3\right)^2\)
\(\Leftrightarrow8\left(m-1\right)^3-3\left(-3\right).2\left(m-1\right)=9\left(m-1\right)\)
\(\Leftrightarrow8\left(m-1\right)^3+9\left(m-1\right)=0\)
\(\Leftrightarrow\left(m-1\right)\left[8\left(m-1\right)^2+9\right]=0\)
\(\Leftrightarrow m=1\)(do \(8\left(m-1\right)^2+9>0\) với mọi m)
Vậy m=1
Vì \(ac< 0\) \(\Rightarrow\) Phương trình luôn có 2 nghiệm phân biệt
Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m-2\\x_1x_2=-3\end{matrix}\right.\)
Mặt khác: \(\dfrac{x_1}{x_2^2}+\dfrac{x_2}{x_1^2}=m-1\) \(\Rightarrow\dfrac{\left(x_1+x_2\right)\left(x_1^2+x_2^2-x_1x_2\right)}{x_1^2x_2^2}=m-1\)
\(\Leftrightarrow\dfrac{\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-3x_1x_2\right]}{x_1^2x_2^2}=m-1\)
\(\Rightarrow\dfrac{\left(2m-2\right)\left(4m^2-8m+13\right)}{9}=m-1\)
\(\Leftrightarrow...\)
Phương trình có nghiệm \(\Leftrightarrow\Delta'\ge0\Leftrightarrow1-m\ge0\Leftrightarrow m\le1\)
Theo hệ thức Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m\end{matrix}\right.\) (1)
Ta có: \(\dfrac{1}{x^2}+\dfrac{1}{x^2}=1\Leftrightarrow\dfrac{x^2_1+x^2_2}{x^2_1x^2_2}=1\Leftrightarrow\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{\left(x_1x_2\right)^2}=1\) (2)
Từ (1) và (2) \(\Rightarrow4-2m=m^2\Leftrightarrow m^2+2m-4=0\)
\(\Delta'=1+4=5\Rightarrow\sqrt{\Delta'}=\sqrt{5}\Rightarrow\left[{}\begin{matrix}m=-1+\sqrt{5}\left(\text{loại}\right)\\m=-1-\sqrt{5}\left(\text{nhận}\right)\end{matrix}\right.\)
Vậy \(m=-1-\sqrt{5}\)
\(\text{Δ}=\left(-4n\right)^2-4\left(12n-9\right)\)
\(=16n^2-48n+36\)
\(=\left(4n-6\right)^2\)>=0
=>Phương trình luôn có hai nghiệm
Theo đề, ta có: \(2x_1x_2+3\left(x_1+x_2\right)-54=0\)
\(\Leftrightarrow2\left(12n-9\right)+3\cdot4n-54=0\)
=>24n-18+12n-54=0
=>36n-72=0
hay n=2
Có\(\Delta=4\left(m+1\right)^2-4\left(2m-3\right)=4m^2+16>0\forall m\)
=> pt luôn có hai nghiệm pb
Theo viet có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m-3\end{matrix}\right.\)
Có :\(P^2=\left(\dfrac{x_1+x_2}{x_1-x_2}\right)^2=\dfrac{4\left(m+1\right)^2}{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(=\dfrac{4\left(m+1\right)^2}{4\left(m+1\right)^2-4\left(2m-3\right)}=\dfrac{4\left(m+1\right)^2}{4m^2+16}\)\(\ge0\)
\(\Rightarrow P\ge0\)
Dấu = xảy ra khi m=-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
\(x^2+3x+m-3=0\)
Ta có \(\Delta=b^2-4ac\)
\(=3^2-4.1.\left(m-3\right)\)
\(=9-4m+12\)
\(=21-4m\)
Đẻ pt có 2 nghiệm \(x_1;x_2\)\(\Leftrightarrow\Delta\ge0\Leftrightarrow21-4m\ge0\)
\(\Leftrightarrow x\le\frac{21}{4}\)
Áp dụng vi-ét ta có
\(\hept{\begin{cases}x_1+x_2=-3\\x_1.x_2=m-3\end{cases}}\)
Ta có \(\frac{x_1}{x_2}+\frac{x_2}{x_1}=5\Leftrightarrow\frac{x_1^2+x_2^2}{x_1.x_2}=5\)
\(\Leftrightarrow x_1^2+x_2^2=5x_1x_2\)
\(\Leftrightarrow x_1^2+x_2^2-5x_1.x_2=0\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2-5x_1x_2=0\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-7x_1x_2=0\)
\(\Leftrightarrow\left(-3\right)^2-7\left(m-3\right)=0\)
\(\Leftrightarrow9-7m+21=0\)
\(\Leftrightarrow30-7m=0\)
\(\Leftrightarrow7m=30\)
\(\Leftrightarrow m=\frac{30}{7}\) (TM)
Vậy \(m=\frac{30}{7}\) thì thỏa mãn bài toán
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