Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\\x_1x_2=\dfrac{c}{a}=-4\end{matrix}\right.\)
a: \(A=x_1^2+x_2^2-x_1x_2\)
\(=\left(x_1+x_2\right)^2-3x_1x_2=2^2-3\cdot\left(-4\right)=4+12=16\)
b: \(B=x_1^3+x_2^3-x_1-x_2\)
\(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)-\left(x_1+x_2\right)\)
\(=2^3-3\cdot2\cdot\left(-4\right)-2=8-2+24=6+24=30\)
c: \(C=\left(x_1+2\right)\left(x_2+2\right)=x_1x_2+2\left(x_1+x_2\right)+4\)
\(=-4+2\cdot2+4=4\)
d: \(D=\left(x_1+1\right)^2+\left(x_2+1\right)^2\)
\(=x_1^2+x_2^2+2\left(x_1+x_2\right)+2\)
\(=\left(x_1+x_2\right)^2-2x_1x_2+2\left(x_1+x_2\right)+2\)
\(=2^2-2\cdot\left(-4\right)+2\cdot2+2\)
=4+8+4+2
=8+10
=18
e: \(E=\dfrac{x_1^2}{x_2}+\dfrac{x_2^2}{x_1}=\dfrac{x_1^3+x_2^3}{x_1x_2}=\dfrac{\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)}{x_1x_2}\)
\(=\dfrac{2^3-3\cdot\left(-4\right)\cdot2}{-4}=\dfrac{8+3\cdot4\cdot2}{4}=2+3\cdot2=2+6=8\)
f: \(F=\dfrac{1}{x_1-2}+\dfrac{1}{x_2-2}=\dfrac{x_2-2+x_1-2}{\left(x_1-2\right)\left(x_2-2\right)}\)
\(=\dfrac{x_1+x_2-4}{x_1x_2-2\left(x_1+x_2\right)+4}=\dfrac{2-4}{-4-2\cdot2+4}=\dfrac{-2}{-4}=\dfrac{1}{2}\)
