a: Sửa đề: \(x^2-5x+4=0\)
\(x_1+x_2=-\dfrac{b}{a}=5;x_1x_2=\dfrac{c}{a}=4\)
\(\left(x_1+\dfrac{1}{x_2}\right)\left(x_2+\dfrac{1}{x_1}\right)\)
\(=\dfrac{x_1x_2+1}{x_2}\cdot\dfrac{x_2x_1+1}{x_1}\)
\(=\dfrac{4+1}{x_1x_2}\cdot\left(4+1\right)=\dfrac{25}{4}\)
b: \(\left|\dfrac{2}{x_2^3}-\dfrac{2}{x_1^3}\right|=\left|2\cdot\dfrac{x_1^3-x_2^3}{\left(x_1x_2\right)^3}\right|\)
\(=\left|\dfrac{2}{4^3}\cdot\left(x_1-x_2\right)\left(x_1^2+x_2^2+x_1x_2\right)\right|\)
\(=\dfrac{1}{32}\cdot\left|\left(x_1-x_2\right)\right|\cdot\left|\left(x_1+x_2\right)^2-x_1x_2\right|\)
\(=\dfrac{1}{32}\cdot\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\cdot\left|5^2-4\right|\)
\(=\dfrac{21}{32}\cdot\sqrt{5^2-4\cdot4}=\dfrac{21}{32}\cdot3=\dfrac{63}{32}\)
