Question

Срочно нужно! Распишите как вы решали
$left { {{x^2+y^2=17} atop {5x-3y=17}} right.$

Answer 1

$left { {{x^2+y^2=17} atop {5x-3y=17}} right. left { {{frac{9y^2+102y+289}{25}+y^2=17} atop {x=frac{3y+17}{5}}} right. left { {{9y^2+102y+289+25y^2=425} atop {x=0,6y+3,4}} right. \34y^2+102y-136=0\y^2+3y-4=0\D=9+16=25=5^2\y_{1,2}=frac{-3б5}{2} =left | {{1} atop {-4}} right. \ left { {{y_1=1} atop {x_1=0,6+3,4}} right. left { {{y_1=1} atop {x_1=4}} right. left { {{y_2=-4} atop {x_2=-2,4+3,4}} right. left { {{y_2=-4} atop {x_2=1}} right.$

Ответ: $(4;1);(1;-4)$

Answer 2

Спасибки.