a)Có ac=-1<0

=>pt luôn có hai nghiệm trái dấu

b)Do x₁;x₂ là hai nghiệm của pt

=> \(\left\{{}\begin{matrix}x_1^2-mx_1-1=0\\x_2^2-mx_2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_1^2-1=mx_1\\x_2^2-1=mx_2\end{matrix}\right.\)

=>\(P=\dfrac{mx_1+x_1}{x_1}-\dfrac{mx_2+x_2}{x_2}\)\(=m+1-\left(m+1\right)=0\)

\(\Delta=b^2-4ac=m^2+16\)

=> Pt luôn có 2 nghiệm phân biệt

Theo Vi-ét : \(\hept{\begin{cases}x_1+x_2=-\frac{b}{a}=m\\x_1\cdot x_2=\frac{c}{a}=-4\end{cases}}\)

Thay vào A ta được : \(A=\frac{2m+7}{m^2+8}\) 

=> Min A = -1/8 khi m=-8

giải thích đi cậu sao ra được min vậy

a: x1+x2=-2; x1x2=-4

x1+x2+2+2=-2+2+2=2

(x1+2)(x2+2)=x1x2+2(x1+x2)+4

=-4+2*(-2)+4=-4

Phương trình cần tìm là x^2-2x-4=0

b: \(\dfrac{1}{x_1+1}+\dfrac{1}{x_2+1}=\dfrac{x_1+x_2+2}{\left(x_1+1\right)\left(x_2+1\right)}\)

\(=\dfrac{x_1+x_2+2}{x_1x_2+\left(x_1+x_2\right)+1}\)

\(=\dfrac{-2+2}{-4+\left(-2\right)+1}=0\)

\(\dfrac{1}{x_1+1}\cdot\dfrac{1}{x_2+1}=\dfrac{1}{x_1x_2+x_1+x_2+1}=\dfrac{1}{-4-2+1}=\dfrac{-1}{5}\)

Phương trình cần tìm sẽ là; x^2-1/5=0

c: \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=\dfrac{x_1^2+x_2^2}{x_1x_2}=\dfrac{\left(-2\right)^2-2\cdot\left(-4\right)}{-4}=\dfrac{4+8}{-4}=-3\)

x1/x2*x2/x1=1

Phương trình cần tìm sẽ là:

x^2+3x+1=0

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

\(A=\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{-1}{-2+\sqrt{2}}=\dfrac{2+\sqrt{2}}{2}\)

\(B=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=\left(-1\right)^2-2\left(-2+\sqrt{2}\right)=5-2\sqrt{2}\)

1. Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{4}{3}\\x_1.x_2=\dfrac{1}{3}\end{matrix}\right.\)

\(C=\dfrac{x_1}{x_2-1}+\dfrac{x_2}{x_1-1}=\dfrac{x_1\left(x_1-1\right)+x_2\left(x_2-1\right)}{\left(x_1-1\right)\left(x_2-1\right)}\)

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

  \(=\dfrac{\left(-\dfrac{4}{3}\right)^2-2.\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)}{\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)+1}=\dfrac{\dfrac{22}{9}}{\dfrac{8}{3}}=\dfrac{11}{12}\)

\(1,3x^2+4x+1=0\)

Do pt có 2 nghiệm \(x_1,x_2\) nên theo đ/l Vi-ét ta có :

\(\left\{{}\begin{matrix}S=x_1+x_2=\dfrac{-b}{a}=-\dfrac{4}{3}\\P=x_1x_2=\dfrac{c}{a}=\dfrac{1}{3}\end{matrix}\right.\)

Ta có :

\(C=\dfrac{x_1}{x_2-1}+\dfrac{x_2}{x_1-1}\)

\(=\dfrac{x_1\left(x_1-1\right)+x_2\left(x_2-1\right)}{\left(x_2-1\right)\left(x_1-1\right)}\)

\(=\dfrac{x_1^2-x_1+x_2^2-x_2}{x_1x_2-x_2-x_1+1}\)

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

\(=\dfrac{S^2-2P-S}{P-S+1}\)

\(=\dfrac{\left(-\dfrac{4}{3}\right)^2-2.\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)}{\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)+1}\)

\(=\dfrac{11}{12}\)

Vậy \(C=\dfrac{11}{12}\)

\(3,3x^2-7x-1=0\)

Do pt có 2 nghiệm \(x_1,x_2\) nên theo đ/l Vi-ét ta có :

\(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=\dfrac{7}{3}\\P=x_1x_2=\dfrac{c}{a}=-\dfrac{1}{3}\end{matrix}\right.\)

Ta có :

\(B=\dfrac{2x_2^2}{x_1+x_2}+2x_1\)

\(=\dfrac{2x_2^2+2x_1\left(x_1+x_2\right)}{x_1+x_2}\)

\(=\dfrac{2x_2^2+2x_1^2+2x_1x_2}{x_1+x_2}\)

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

\(=\dfrac{2\left(S^2-2P\right)+2P}{S}\)

\(=\dfrac{2\left(\dfrac{7}{3}^2-2\left(-\dfrac{1}{3}\right)\right)+2\left(-\dfrac{1}{3}\right)}{\dfrac{7}{3}}\)

\(=\dfrac{104}{21}\)

Vậy \(B=\dfrac{104}{21}\)

a: A=x1+x2=-5/2

b: \(=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{-5}{2}:\left(-1\right)=\dfrac{5}{2}\)

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

\(=\left(-\dfrac{5}{2}\right)^3-3\cdot\dfrac{-5}{2}\cdot\left(-1\right)\)

\(=-\dfrac{125}{8}-\dfrac{15}{2}=\dfrac{-185}{8}\)

e: \(E=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)

\(=\sqrt{\left(-\dfrac{5}{2}\right)^2-4\cdot\left(-1\right)}=\sqrt{\dfrac{25}{4}+4}=\dfrac{\sqrt{41}}{2}\)