a)Có ac=-1<0

=>pt luôn có hai nghiệm trái dấu

b)Do x₁;x₂ là hai nghiệm của pt

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

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

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

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

Δ=(m+2)^2-4*2m=(m-2)^2

Để PT có hai nghiệm pb thì m-2<>0

=>m<>2

\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1x_2}{4}\)

=>\(\dfrac{x_1+x_2}{x_1x_2}=\dfrac{x_1x_2}{4}\)

=>\(\dfrac{m+2}{2m}=\dfrac{2m}{4}=\dfrac{m}{2}\)

=>2m^2=2m+4

=>m^2-m-2=0

=>m=2(loại) hoặc m=-1

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:

Ta có: \(\Delta=\left[-2\left(m-1\right)\right]^2-4\cdot1\cdot\left(m+1\right)\)

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

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

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

Để phương trình có nghiệm thì \(\text{Δ}\ge0\)

\(\Leftrightarrow4m^2-12m\ge0\)

\(\Leftrightarrow4m\left(m-3\right)\ge0\)

\(\Leftrightarrow m\left(m-3\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}m\ge3\\m\le0\end{matrix}\right.\)

Khi \(\left[{}\begin{matrix}m\ge3\\m\le0\end{matrix}\right.\), Áp dụng hệ thức Vi-et, ta có:

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

Ta có: \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=4\)

\(\Leftrightarrow\dfrac{x_1^2+x_2^2}{x_1\cdot x_2}=4\)

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

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

\(\Leftrightarrow4m^2-8m+4-2m-2=4\left(m+1\right)\)

\(\Leftrightarrow4m^2-10m+2-4m-4=0\)

\(\Leftrightarrow4m^2-14m-2=0\)

Đến đây bạn tự làm nhé, chỉ cần tìm m và đối chiều với điều kiện thôi

Pt có 2 nghiệm

\(\to \Delta=[-2(m-1)]^2-4.1.(m+1)=4m^2-8m+4-4m-4=4m^2-12m\ge 0\)

\(\leftrightarrow m^2-3m\ge 0\)

\(\leftrightarrow m(m-3)\ge 0\)

\(\leftrightarrow \begin{cases}m\ge 0\\m-3\ge 0\end{cases}\quad or\quad \begin{cases}m\le 0\\m-3\le 0\end{cases}\)

\(\leftrightarrow m\ge 3\quad or\quad m\le 0\)

Theo Viét

\(\begin{cases}x_1+x_2=2(m-1)\\x_1x_2=m+1\end{cases}\)

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

\(\leftrightarrow \dfrac{x_1^2+x_2^2}{x_1x_2}=4\)

\(\leftrightarrow \dfrac{(x_1+x_2)^2-2x_1x_2}{x_1x_2}=4\)

\(\leftrightarrow \dfrac{[2(m-1)]^2-2.(m+1)}{m+1}=4\)

\(\leftrightarrow 4m^2-8m+4-2m-2=4(m+1)\)

\(\leftrightarrow 4m^2-10m+2-4m-4=0\)

\(\leftrightarrow 4m^2-14m-2=0\)

\(\leftrightarrow 2m^2-7m-1=0 (*)\)

\(\Delta_{*}=(-7)^2-4.2.(-1)=49+8=57>0\)

\(\to\) Pt (*) có 2 nghiệm phân biệt

\(m_1=\dfrac{7+\sqrt{57}}{2}(TM)\)

\(m_2=\dfrac{7-\sqrt{57}}{2}(TM)\)

Vậy \(m=\dfrac{7\pm \sqrt{57}}{2}\) thỏa mãn hệ thức

a) Ta có: \(\text{Δ}=\left[-2\left(m-1\right)\right]^2-4\cdot1\cdot\left(-m\right)\)

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

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

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

\(=4m^2-4m+1+3\)

\(=\left(2m-1\right)^2+3>0\forall x\)

Do đó: Phương trình luôn có hai nghiệm x1,x2 với mọi m(Đpcm)

b) Áp dụng hệ thức Vi-et, ta được:

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

Ta có: \(y_1+y_2=x_1+\dfrac{1}{x_2}+x_2+\dfrac{1}{x_1}\)

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

\(=\left(2m-2\right)+\dfrac{2m-2}{-m}\)

\(=2m-2-\dfrac{2m-2}{m}\)

\(=\dfrac{2m^2-2m-2m+2}{m}\)

\(=\dfrac{2m^2-4m+2}{m}\)

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

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

Ta có: \(y_1y_2=\left(x_1+\dfrac{1}{x_2}\right)\left(x_2+\dfrac{1}{x_1}\right)\)

\(=x_1x_2+2+\dfrac{1}{x_1x_2}\)

\(=-m+2+\dfrac{1}{-m}\)

\(=-m+2-\dfrac{1}{m}\)

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

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

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

Phương trình đó sẽ là:

\(x^2-\dfrac{2\left(m-1\right)^2}{m}x-\dfrac{\left(m-1\right)^2}{m}=0\)

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 \(B=\dfrac{104}{21}\)