Х1^4+х2^4; х1/х2+х2/х1 корни уравнения 2х^2-11х+13=0

Решите задачу:

$\mathtt{2x^2-11x+13=0;~x^2-5,5x+6,5=0,~\to~\left[\begin{array}{ccc}\mathtt{x_1+x_2=5,5}\\\mathtt{x_1x_2=6,5}\end{array}\right}$

$\mathtt{x_1^4+x_2^4=(x_1^2)^2+(x_2^2)^2=(x_1^2+x_2^2)^2-2x_1^2x_2^2=}\\\mathtt{([x_1+x_2]^2-2x_1x_2)^2-2x_1^2x_2^2=(5,5^2-2*6,5)^2-2*6,5^2=}\\\mathtt{17,25^2-84,5=213,0625=213\frac{1}{16}}$

$\mathtt{\frac{x_1}{x_2}+\frac{x_2}{x_1}=\frac{x_1^2+x_2^2}{x_1x_2}=\frac{(x_1+x_2)^2-2x_1x_2}{x_1x_2}=\frac{(x_1+x_2)^2}{x_1x_2}-2=\frac{5,5^2}{6,5}-2=}\\\\\mathtt{\frac{5,5*55}{65}-2=\frac{5,5*11}{13}-2=\frac{60,5-26}{13}=\frac{34,5}{13}=\frac{345}{130}=\frac{69}{26}=2\frac{17}{26}}$