Theo Vi-ét, ta có : \(\left\{{}\begin{matrix}x_!+x_2=-\dfrac{b}{a}=3\\x_1x_2=\dfrac{c}{a}=-7\end{matrix}\right.\)
\(1)A=\dfrac{1}{x_1-1}+\dfrac{1}{x_2-1}\\ =\dfrac{x_2-1+x_1-1}{\left(x_1-1\right)\left(x_2-1\right)}\\ =\dfrac{x_1+x_2-2}{x_1x_2-\left(x_1+x_2\right)+1}\\ =\dfrac{3-2}{-7-3+1}\\ =-\dfrac{1}{9}\)
\(2)B=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=3^2-2.\left(-7\right)=23\)
\(3)C=x_1^3+x_2^3\\ =\left(x_1+x_2\right)\left(x_1^2+x_1x_2+x_2^2\right)\\ =\left(x_1+x_2\right)\left[x_1x_2+\left(x_1+x_2\right)^2-2x_1x_2\right]\\ =3\left(-7+3^2-2.\left(-7\right)\right)\\ =48\)
\(4)D=x_1^4+x_2^4\\ =\left(x_1^2+x_2^2\right)^2-2\left(x_1x_2\right)^2\\ =\left[\left(x_1+x_2\right)^2-2x_1x_2\right]^2-2\left(x_1x_2\right)^2\\ =\left(3^2+14\right)^2-2\left(-7\right)^2\\ =431\)
\(5)E=\left(3x_1+x_2\right)\left(x_1+3x_2\right)\\ =3x^2_1+9x_1x_2+x_1x_2+3x_2^2\\ =3\left(x_1^2+x_2^2\right)+10x_1x_2\\ =3.23+10.\left(-7\right)\\ =-1\)
1: \(A=\dfrac{x_1-1+x_2-1}{\left(x_1-1\right)\left(x_2-1\right)}\)
\(=\dfrac{x_1+x_2-2}{x_1x_2-\left(x_1+x_2\right)-1}\)
\(=\dfrac{3-2}{-7-3-1}=\dfrac{1}{-11}=\dfrac{-1}{11}\)
2: B=(x1+x2)^2-2x1x2
=3^2-2*(-7)=9+14=23
3: C=(x1+x2)^3-3x1x2(x1+x2)
=3^2-3*3*(-7)
=9+9*7=72
4: D=(x1^2+x2^2)^2-2*(x1x2)^2
=23^2-2*(-7)^2=431
5: E=3x1^2+9x1x2+x1x2+3x2^2
=3(x1^2+x2^2)+10x1x2
=3*23+10*(-7)=-1
