PT: \(x^2-13x-7=0\)
Theo vi-ét ta có: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left(-13\right)}{1}=13\\x_1x_2=\dfrac{c}{a}=\dfrac{-7}{1}=-7\end{matrix}\right.\)
a) \(x_1^3+x^3_2=\left(x_1+x_2\right)\left(x^2_1-x_1x_2+x^2_2\right)\)
\(=\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-2x_1x_2-x_1x_2\right]\)
\(=\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-3x_1x_2\right]\)
\(=13\cdot\left(13^2-3\cdot-7\right)=2470\)
b) \(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{13}{-7}\)
c) \(\dfrac{x_1}{x^2_2}+\dfrac{x_2}{x^2_1}=\dfrac{x_1^3+x^3_2}{x^2_1x^2_2}=\dfrac{2470}{\left(-7\right)^2}=\dfrac{2470}{49}\)
Nhãn
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