\(A=\dfrac{\left(x_1+x_2\right)^2+3x_1x_2}{4x_1x_2\left(x_1+x_2\right)}=\dfrac{9+3}{4\cdot1\left(-3\right)}=\dfrac{12}{-12}=-1\)

\(\Delta'=m-1\ge0\Rightarrow m\ge1\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=m^2-m+1\end{matrix}\right.\)

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

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

\(=8m^3-3.2m\left(m^2-m+1\right)-4m\)

\(=2m^3+6m^2-10m\)

\(=2\left(m^3+3m^2-5m+1\right)-2\)

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

Do \(m\ge1\Rightarrow\left\{{}\begin{matrix}m-1\ge0\\\left(m^2-1\right)+4m>0\end{matrix}\right.\)

\(\Rightarrow2\left(m-1\right)\left[\left(m^2-1\right)+4m\right]\ge0\)

\(\Rightarrow A\ge-2\)

\(A_{min}=-2\) khi \(m=1\)

\(\Delta'=\left(m+1\right)^2-\left(2m+10\right)=m^2-9\)

Pt có 2 nghiệm khi \(m^2-9\ge0\Rightarrow\left[{}\begin{matrix}m\ge3\\m\le-3\end{matrix}\right.\)

Khi đó theo định lý Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m+10\end{matrix}\right.\)

\(A=x_1^2+x_2^2+14x_1x_2=\left(x_1+x_2\right)^2+12x_1x_2\)

\(=4\left(m+1\right)^2+12\left(2m+10\right)\)

\(=4\left(m+4\right)^2+60\ge60\)

Dấu "=" xảy ra khi \(m+4=0\Rightarrow m=-4\) (thỏa mãn)

a) ∆' = [-(m - 3)]² - (m² + 3)

= m² - 6m + 9 - m² - 3

= -6m + 6

Để phương trình đã cho có 2 nghiệm thì ∆' ≥ 0

⇔ -6m + 6 ≥ 0

⇔ 6m ≤ 6

⇔ m ≤ 1

Vậy m ≤ 1 thì phương trình đã cho luôn có 2 nghiệm

b) Theo định lý Viét, ta có:

x₁ + x₂ = 2(m - 3) = 2m - 6

x₁x₂ = m² + 3

Ta có:

(x₁ - x₂)² - 5x₁x₂ = 4

⇔ x₁² - 2x₁x₂ + x₂² - 5x₁x₂ = 4

⇔ x₁² + 2x₁x₂ + x₂² - 2x₁x₂ - 2x₁x₂ - 5x₁x₂ = 4

⇔ (x₁ + x₂)² - 9x₁x₂ = 4

⇔ (2m - 6)² - 9(m² + 3) = 4

⇔ 4m² - 24m + 36 - 9m² - 27 = 4

⇔ -5m² - 24m + 9 = 4

⇔ 5m² + 24m - 5 = 0

⇔ 5m² + 25m - m - 5 = 0

⇔ (5m² + 25m) - (m + 5) = 0

⇔ 5m(m + 5) - (m + 5) = 0

⇔ (m + 5)(5m - 1) = 0

⇔ m + 5 = 0 hoặc 5m - 1 = 0

*) m + 5 = 0

⇔ m = -5 (nhận)

*) 5m - 1 = 0

⇔ m = 1/5 (nhận)

Vậy m = -5; m = 1/5 thì phương trình đã cho có 2 nghiệm thỏa mãn yêu cầu

a: \(\Delta=\left[-2\left(m-3\right)\right]^2-4\cdot1\cdot\left(m^2+3\right)\)

\(=\left(2m-6\right)^2-4\left(m^2+3\right)\)

\(=4m^2-24m+36-4m^2-12=-24m+24\)

Để phương trình có hai nghiệm thì \(\Delta>=0\)

=>-24m+24>=0

=>-24m>=-24

=>m<=1

b: Theo Vi-et, ta có:

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

\(\left(x_1-x_2\right)^2-5x_1x_2=4\)

=>\(\left(x_1+x_2\right)^2-4x_1x_2-5x_2x_1=4\)

=>\(\left(x_1+x_2\right)^2-9x_1x_2=4\)

=>\(\left(2m-6\right)^2-9\left(m^2+3\right)=4\)

=>\(4m^2-24m+36-9m^2-27-4=0\)

=>\(-5m^2-24m+5=0\)

=>\(-5m^2-25m+m+5=0\)

=>\(-5m\left(m+5\right)+\left(m+5\right)=0\)

=>(m+5)(-5m+1)=0

=>\(\left[{}\begin{matrix}m+5=0\\-5m+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=-5\left(nhận\right)\\m=\dfrac{1}{5}\left(nhận\right)\end{matrix}\right.\)

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

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

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

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

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

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

   \(=4m^2-8m+8+8\)

   \(=4m^2-8m+16\)

   \(=3m^2+\left(m-4\right)^2\)

Để pt có 2 nghiệm phân biệt thì \(\Delta>0\)

                                                  \(\Leftrightarrow\left\{{}\begin{matrix}m>0\\m>4\end{matrix}\right.\) \(\rightarrow m>4\)

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

\(A=x_1^2+4x_2^2\)

\(A=x_1^2+\left(2x_2\right)^2\)

\(\Rightarrow Min_A=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=0\\x_2=0\end{matrix}\right.\)

Thế vào (1) ta được: \(0=2m-2\)

                                \(\Leftrightarrow m=1\)