a. Theo định lý Viet: \(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=-2\end{matrix}\right.\)
\(A=x_1^2+3x_1x_2+2x_2^2-x_2^2=x_1^2+3x_1x_2+x_2^2\)
\(=\left(x_1+x_2\right)^2+x_1x_2=5^2-2=23\)
\(B=x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=5^3-3.\left(-2\right).5=155\)
\(C=\dfrac{x_1+1}{x_2}+\dfrac{x_2+1}{x_1}=\dfrac{x_1^2+x_2^2+x_1+x_2}{x_1x_2}=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2+x_1+x_2}{x_1x_2}\)
\(=\dfrac{5^2-2.\left(-2\right)+5}{-2}=-17\)
b.
Theo Viet đảo, a và b là nghiệm của pt:
\(x^2-5x+6=0\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy \(\left(a;b\right)=\left(2;3\right);\left(3;2\right)\)
\(3.\)
Ta có: \(\Delta=\left(-5\right)^2-4.1.\left(-2\right)=33>0\)
ĐL Vi - et:
\(\left[{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}=\dfrac{5}{1}=5\\x_1.x_2=\dfrac{c}{a}=-\dfrac{2}{1}=-2\end{matrix}\right.\)
Theo đề bài ta có:
\(A=\left(x_1+x_2\right)\left(x_1+2x_2\right)-x_2^2=x_1^2+2x_1x_2+x_2^2+2x_2^2-x_2^2=\left(x_1+x_2\right)^2+x_2^2=5^2+x_2^2=25+x_2^2\)
\(P=x^3_1+x^3_2=x^3_1+x^3_2+3x_1^2x_2+3x_1x_2^2-3x_1^2x_2-3x_1x_2^2=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=5^3-3.\left(-2\right).5=155\)
\(C=\dfrac{x_1+1}{x_2}+\dfrac{x_2+1}{x_1}\)
\(=\dfrac{x_2\left(x_1+1\right)}{x_1x_2}+\dfrac{x_1\left(x_2+1\right)}{x_1x_2}=x_2\left(x_1+1\right)+x_1\left(x_2+1\right)\)
\(=x_1x_2+x_2+x_1x_2+x_1=x_1x_2+x_1x_2+\left(x_1+x_2\right)=\left(-2\right)+\left(-2\right)+5=1\)
