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

Để pt có 2 nghiệm thì \(\Delta'\ge0\Leftrightarrow2-m\ge0\Leftrightarrow m\le2\)

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m-1\end{matrix}\right.\)

\(x^4_1-x^3_1=x^4_2-x^3_2\\ \Leftrightarrow\left(x^4_1-x_2^4\right)-\left(x^3_1+x^3_2\right)=0\\ \Leftrightarrow\left(x^2_1-x^2_2\right)\left(x^2_1+x^2_2\right)-\left(x_1+x_2\right)\left(x^2_1+x^2_2-x_1x_2\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-3x_1x_2\right]=0\\ \Leftrightarrow\left(m-1\right).2\left[2^2-2\left(m-1\right)\right]-2\left[2^2-3\left(m-1\right)\right]=0\)

\(\Leftrightarrow\left(2m-2\right)\left(4-2m+2\right)-2\left(4-3m+3\right)=0\)

\(\Leftrightarrow\left(2m-2\right)\left(6-2m\right)-2\left(7-3m\right)=0\)

\(\Leftrightarrow...\)

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

Để pt có 2 nghiệm thì \(\Delta'\ge0\Leftrightarrow2-m\ge0\Leftrightarrow m\le2\)

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m-1\end{matrix}\right.\)

\(x^4_1-x^3_1=x^4_2-x^3_2\\ \Leftrightarrow\left(x^4_1-x_2^4\right)-\left(x^3_1-x^3_2\right)=0\\ \Leftrightarrow\left(x^2_1-x^2_2\right)\left(x^2_1+x^2_2\right)-\left(x_1-x_2\right)\left(x^2_1+x^2_2+x_1x_2\right)=0\)

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

\(\Leftrightarrow\left(x_1-x_2\right).2\left(4-2m+2\right)-\left(x_1-x_2\right)\left(4-m+1\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right).2\left(6-2m\right)-\left(x_1-x_2\right)\left(5-m\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(12-4m-5+m\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(7-3m\right)=0\)

\(\Leftrightarrow...\)

a, \(\Delta=m^2-4\left(-4\right)=m^2+16\)> 0 

Vậy pt luôn có 2 nghiệm pb 

b, Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=-4\end{matrix}\right.\)

Ta có \(\left(x_1+x_2\right)^2-2x_1x_2=5\)

Thay vào ta được \(m^2-2\left(-4\right)=5\Leftrightarrow m^2+3=0\left(voli\right)\)

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

\(\dfrac{\left(x_1^2-2\right)\left(x_2^2-2\right)}{\left(x_1-1\right)\left(x_2-1\right)}=4\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1^2+x_2^2\right)+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)

\(\Rightarrow\dfrac{\left(x_1x_2\right)^2-2\left(x_1+x_2\right)^2+4x_1x_2+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)

\(\Rightarrow\dfrac{\left(m-2\right)^2-2m^2+4\left(m-2\right)+4}{m-2-m+1}=4\)

\(\Rightarrow-m^2=-4\Rightarrow m=\pm2\)

a, x = 3 , x= -1

b, m = 3 , m = 1

Ta có \(ac=-m^2-2< 0\) ; \(\forall m\) nên pt đã cho luôn có 2 nghiệm trái dấu

Mà \(x_1< x_2\Rightarrow\left\{{}\begin{matrix}x_1< 0\\x_2>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|x_1\right|=-x_1\\\left|x_2\right|=x_2\end{matrix}\right.\)

\(\Rightarrow2\left|x_1\right|-\left|x_2\right|=4\Leftrightarrow-2x_1-x_2=4\)

Kết hợp với hệ thức Viet: \(x_1+x_2=-m+1\)

\(\Rightarrow\left\{{}\begin{matrix}-2x_1-x_2=4\\x_1+x_2=-m+1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-x_1=-m+5\\x_1+x_2=-m+1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=m-5\\x_2=-2m+6\end{matrix}\right.\)

Thay vào \(x_1x_2=-m^2-2\)

\(\Rightarrow\left(m-5\right)\left(-2m+6\right)=-m^2-2\)

\(\Leftrightarrow m^2-16m+28=0\Rightarrow\left[{}\begin{matrix}m=2\\m=14\end{matrix}\right.\)

Cách ngắn ngọn nhất:

⇔[x=2x=2m⇔[�=2�=2�

Phương trình (1) có 2 nghiệm là . Mặt khác phương trình (1) cũng có 2 nghiệm là x₁, x₂ nên ta chia làm 2 trường hợp:

TH₁: .

Có 

TH₂: 

Có 

Vậy m=0 hay m=3

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

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

\(\text{Δ}=2^2-4\cdot1\cdot m=4-4m\)

Để phương trình có hai nghiệm thì Δ>=0

=>-4m+4>=0

=>-4m>=-4

=>m<=1(1)

Theo Vi-et, ta có: 

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

\(\dfrac{x_1^2-3x_1+m}{x_2}+\dfrac{x_2^2-3x_2+m}{x_1}< =2\)

=>\(\dfrac{x_1^3+x_2^3-3\left(x_1^2+x_2^2\right)+m\left(x_1+x_2\right)}{x_1x_2}< =2\)

=>\(\dfrac{\left(x_1+x_2\right)^3-3x_1x_2-3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+m\left(x_1+x_2\right)}{x_1x_2}< =2\)

=>\(\dfrac{\left(-2\right)^3-3\cdot m-3\left[\left(-2\right)^2-2m\right]+m\cdot\left(-2\right)}{m}< =2\)

=>\(\dfrac{-8-3m-3\left(4-2m\right)-2m}{m}-2< =0\)

=>\(\dfrac{-5m-8-12+6m}{m}-2< =0\)

=>\(\dfrac{m-20-2m}{m}< =0\)

=>\(\dfrac{-m-20}{m}< =0\)

=>\(\dfrac{m+20}{m}>=0\)

=>\(\left[{}\begin{matrix}m>0\\m< =-20\end{matrix}\right.\)

Kết hợp (1), ta được: \(\left[{}\begin{matrix}0< m< =1\\m< =-20\end{matrix}\right.\)