\(3x^2+x-2=0\)
\(\Delta=b^2-4ac=1-4.3.\left(-2\right)=25>0\)
\(\Rightarrow\) Pt có 2 nghiệm \(x_1,x_2\)
\(a,\) Theo Vi-ét, ta có : \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-\dfrac{1}{3}\\x_1x_2=\dfrac{c}{a}=-\dfrac{2}{3}\end{matrix}\right.\)
\(\left(x_1-x_2\right)^2=x_1^2+x_2^2-2x_1x_2=\left(x_1+x_2\right)^2-2x_1x_2-2x_1x_2\)
\(=\left(x_1+x_2\right)^2-4x_1x_2=\left(-\dfrac{1}{3}\right)^2-4.\left(-\dfrac{2}{3}\right)=\dfrac{25}{9}\)
\(x_1^3-x_2^3=\sqrt{\left(x_1-x_2\right)^2}\left(x_1^2+x_2^2+x_1x_2\right)=\sqrt{\dfrac{25}{9}}\left[\left(x_1+x_2\right)^2-x_1x_2\right]\)
\(=\dfrac{5}{3}.\left[\left(-\dfrac{1}{3}\right)^2+\dfrac{2}{3}\right]=\dfrac{5}{3}.\dfrac{7}{9}=\dfrac{35}{27}\)
\(b,A=3x_1\left(x_1-x_2\right)+3x_2\left(x_2-x_1\right)\\ =3x_1\left(x_1-x_2\right)-3x_2\left(x_1-x_2\right)\\ =3\left(x_1-x_2\right)^2=3.\dfrac{25}{9}\\ =\dfrac{25}{3}\)
\(c,\dfrac{2021}{x_1}+\dfrac{2021}{x_2}=\dfrac{2021\left(x_1+x_2\right)}{x_1x_2}=\dfrac{2021.\left(-\dfrac{1}{3}\right)}{-\dfrac{2}{3}}=\dfrac{2021}{2}\)
\(\dfrac{2x_1-1}{x_2}+\dfrac{2x_2-1}{x_1}\\ =\dfrac{\left(2x_1-1\right)x_1+\left(2x_2-1\right)x_2}{x_1x_2}\\ =\dfrac{2x_1^2-x_1+2x_2^2-x_2}{x_1x_2}\\ =\dfrac{2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-\left(x_1+x_2\right)}{x_1x_2}\)
Tới đây thay Vi-ét ở trên vào.
\(\dfrac{3}{x_1+2}+\dfrac{3}{x_2+2}=\dfrac{3\left(x_1+2+x_2+2\right)}{\left(x_1+2\right)\left(x_2+2\right)}=\dfrac{3\left(x_1+x_2+4\right)}{x_1x_2+2\left(x_1+x_2\right)+4}\)
Tới đây thay Vi-ét ở trên vào.
\(\dfrac{x_1-3}{x_1}+\dfrac{x_2-3}{x_2}=\dfrac{\left(x_1-3\right)x_2+\left(x_2-3\right)x_1}{x_1x_2}=\dfrac{x_1x_2-3x_2+x_1x_2-3x_1}{x_1x_2}=\dfrac{2x_1x_2-3\left(x_1+x_2\right)}{x_1x_2}\)
Tới đây thay Vi-ét ở trên vào.
a: x1*x2=-2/3; x1+x2=-1/3
(x1-x2)^2=(x1+x2)^2-4x1x2=1/9-4*(-2/3)=1/9+8/3=25/9
b: A=3x1^2-3x1x2+3x2^2-3x1x2
=3(x1^2-2x1x2+x2^2)
=3(x1-x2)^2=25/3
c: 2021/x1+2021/x2
=2021(x1+x2)/x1x2
=2021*(-1/3):(-2/3)=4042
