Sửa đề: \(\dfrac{x_1x_2}{x_1+x_2}=-\dfrac{m^2}{2}\)

PT có 2 nghiệm phân biệt \(\Leftrightarrow\Delta>0\)

\(\Leftrightarrow\left(m-3\right)^2+4\left(2m^2-3m\right)>0\\ \Leftrightarrow9m^2-18m+9>0\\ \Leftrightarrow9\left(m-1\right)^2>0\left(\text{luôn đúng},\forall m\ne1\right)\)

Do đó PT có 2 nghiệm phân biệt với mọi \(m\ne1\)

Áp dụng Viét: \(\left\{{}\begin{matrix}x_1+x_2=3-m\\x_1x_2=3m-2m^2\end{matrix}\right.\)

Ta có \(\dfrac{x_1x_2}{x_1+x_2}=-\dfrac{m^2}{2}\Leftrightarrow\dfrac{3m-2m^2}{3-m}=-\dfrac{m^2}{2}\)

\(\Leftrightarrow4m^2-12m=3m^2-m^3\\ \Leftrightarrow m^3+m^2-12m=0\\ \Leftrightarrow m\left(m^2+4m-3m-12\right)=0\\ \Leftrightarrow m\left(m+4\right)\left(m-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=0\\m=-4\\m=3\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}m=0\\m=-4\\m=3\end{matrix}\right.\) thỏa yêu cầu đề

1: \(\text{Δ}=\left(2m-2\right)^2-4\left(m-3\right)\)

\(=4m^2-8m+4-4m+12\)

\(=4m^2-12m+16\)

\(=\left(2m-3\right)^2+7>0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt

2: Theo vi-et, ta được:

\(\left\{{}\begin{matrix}x_1+x_2=2m-2\\x_1x_2=m-3\end{matrix}\right.\)

Ta có: \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=x_1x_2\)

\(\Leftrightarrow x_1^2+x_2^2=\left(m-3\right)^2\)

\(\Leftrightarrow\left(2m-2\right)^2-2\left(m-3\right)-\left(m-3\right)^2=0\)

\(\Leftrightarrow4m^2-16m+4-2m+6-m^2+6m-9=0\)

\(\Leftrightarrow3m^2-12m+1=0\)

\(\text{Δ}=\left(-12\right)^2-4\cdot3\cdot1=144-12=132>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{12-2\sqrt{33}}{6}=\dfrac{6-\sqrt{33}}{3}\\x_2=\dfrac{6+\sqrt{33}}{3}\end{matrix}\right.\)

\(\Delta=\left[-2\left(m+1\right)\right]^2-4\left(m^2-3\right)\)

\(=4m^2+8m+4-4m^2+12=8m+16\)

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, bạn tự làm 

b, \(\Delta'=\left(m+2\right)^2-\left(m^2+m+3\right)=m^2+4m+4-m^2-m-3\)

\(=3m+1\)để pt có 2 nghiệm \(m\ge-\dfrac{1}{3}\)

Ta có \(\dfrac{x_1^2+x_2^2}{x_1x_2}=4\Leftrightarrow\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=4\Rightarrow\left(x_1+x_2\right)^2-6x_1x_2=0\)

\(\Rightarrow4\left(m+2\right)^2-6\left(m^2+m+3\right)=0\)

\(\Leftrightarrow4m^2+16m+16-6m^2-6m-18=0\)

\(\Leftrightarrow-2m^2+10m-2=0\Leftrightarrow m^2-5m+1=0\Leftrightarrow m=\dfrac{5\pm\sqrt{21}}{2}\)(tm) 

a: Thay m=-5 vào (1), ta được:

\(x^2+2\left(-5+1\right)x-5-4=0\)

\(\Leftrightarrow x^2-8x-9=0\)

=>(x-9)(x+1)=0

=>x=9 hoặc x=-1

b: \(\text{Δ}=\left(2m+2\right)^2-4\left(m-4\right)=4m^2+8m+4-4m+16=4m^2+4m+20>0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt 

\(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=-3\)

\(\Leftrightarrow x_1^2+x_2^2=-3x_1x_2\)

\(\Leftrightarrow\left(x_1+x_2\right)^2+x_1x_2=0\)

\(\Leftrightarrow\left(2m+2\right)^2+m-4=0\)

\(\Leftrightarrow4m^2+9m=0\)

=>m(4m+9)=0

=>m=0 hoặc m=-9/4

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: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)\\x_1x_2=\dfrac{c}{a}=m+3\end{matrix}\right.\)

\(\Leftrightarrow\left[-2\left(m-1\right)^2\right]-2\left(m+3\right)=1\)

\(\Leftrightarrow4m^2-8m+4-2m-6-1=0\)

\(\Leftrightarrow4m^2-10m-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m_1=\dfrac{5+\sqrt{37}}{4}\left(ktm\right)\\m_2=\dfrac{5-\sqrt{37}}{4}\left(tm\right)\end{matrix}\right.\Rightarrow m=\dfrac{5-\sqrt{37}}{4}\)

Câu c:

ĐKXĐ: m<>-1

Ta có: \(\Delta=\left[-2\left(m-1\right)\right]^2-4\left(m+1\right)\left(m-2\right)\)

\(=\left(2m-2\right)^2-4\left(m^2-m-2\right)\)

\(=4m^2-8m+4-4m^2+4m-8\)

\(=-4m-4\)

Để phương trình có hai nghiệm phân biệt thì -4m-4>0

hay m<-1

Áp dụng hệ thức Vi-et, ta được:

\(\left\{{}\begin{matrix}x_1\cdot x_2=\dfrac{m-2}{m+1}\\x_1+x_2=\dfrac{2\left(m-1\right)}{m+1}\end{matrix}\right.\)

\(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=4\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=4x_1x_2\)

\(\Leftrightarrow\left(\dfrac{2m-2}{m+1}\right)^2-6\cdot\dfrac{m-2}{m+1}=0\)

\(\Leftrightarrow\left(2m-2\right)^2-6\left(m^2-m-2\right)=0\)

\(\Leftrightarrow4m^2-8m+4-6m^2+6m+12=0\)

\(\Leftrightarrow-2m^2-2m+16=0\)

\(\Leftrightarrow m^2-m-8=0\)

Đến đây bạn tự giải nhé

PT có 2 nghiệm \(\Leftrightarrow\Delta=4\left(m-1\right)^2-4\left(m-2\right)\left(m+1\right)\ge0\)

\(\Leftrightarrow4m^2-8m+4-4m^2+4m+8\ge0\\ \Leftrightarrow12-4m\ge0\\ \Leftrightarrow m\le3\)

Áp dụng Viét: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2\left(m-1\right)}{m+1}\\x_1x_2=\dfrac{m-2}{m+1}\end{matrix}\right.\)

\(\dfrac{x_2}{x_1}+\dfrac{x_1}{x_2}=-4\\ \Leftrightarrow\dfrac{x_1^2+x_2^2}{x_1x_2}=-4\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=-4x_1x_2\\ \Leftrightarrow\left(x_1+x_2\right)^2=-2x_1x_2\\ \Leftrightarrow\dfrac{4\left(m-1\right)^2}{\left(m+1\right)^2}=\dfrac{4-2m}{m+1}\\ \Leftrightarrow4\left(m-1\right)^2=\left(4-2m\right)^2\\ \Leftrightarrow4m^2-8m+4=16-16m+4m^2\\ \Leftrightarrow8m=12\Leftrightarrow m=\dfrac{3}{2}\left(tm\right)\)

(a) Khi \(m=2,\left(1\right)\Leftrightarrow x^2-4x-5=0\left(2\right)\).

Phương trình (2) có \(a-b+c=1-\left(-4\right)+\left(-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{c}{a}=5\end{matrix}\right.\).

Vậy: Khi \(m=2,S=\left\{-1;5\right\}\).

(b) Điều kiện: \(x_1,x_2\ne0\Rightarrow m\in R\)

Phương trình có nghiệm khi:

\(\Delta'=\left(-m\right)^2-1\cdot\left(-m^2-1\right)\ge0\)

\(\Leftrightarrow2m^2+1\ge0\left(LĐ\right)\)

Suy ra, phương trình (1) có nghiệm với mọi \(m\).

Theo định lí Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m\\x_1x_2=\dfrac{c}{a}=-m^2-1\end{matrix}\right.\)

Theo đề: \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=-\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{x_1^2+x_2^2}{x_1x_2}=-\dfrac{5}{2}\Leftrightarrow\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=-\dfrac{5}{2}\)

\(\Leftrightarrow2\left(x_1+x_2\right)^2+x_1x_2=0\)

\(\Leftrightarrow2\left(2m\right)^2+\left(-m^2-1\right)=0\)

\(\Leftrightarrow7m^2=1\Leftrightarrow m=\pm\dfrac{\sqrt{7}}{7}\) (thỏa mãn).

Vậy: \(m=\pm\dfrac{\sqrt{7}}{7}.\)