Question

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

Answer 1

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


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


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