Question

Решите систему уравнении x^2+y^2=26,x-y=4

Answer 1

$displaystyle tt left { {{x^2+y^2=26} atop {x-y=4}} right. to left { {{x^2+y^2=26} atop {x=4+y}} right. to left { {{(4+y)^2+y^2=26} atop {x=4+y}} right. \4^2+2cdot4cdot y+y^2+y^2=26\16+8y+y^2+y^2=26\2y^2+8y+16-26=0\2y^2+8y-10=0 : : : : | div 2\y^2+4y-5=0\D=4^2-4cdot1cdot(-5)=16+20=36\sqrt{D}=sqrt{36}=6\x_1=frac{-4+6}{2}=frac{2}{2}=1\x_2=frac{-4-6}{2}=frac{-10}{2}=-5$

$displaystyle tt left { {{y_1=1} atop {x_1=4+1}} right. to left { {{y_1=1} atop {x_1=5}} right. \left { {{y_2=-5} atop {x_2=4-5}} right. toleft { {{y_2=-5} atop {x_2=-1}} right.$

Ответ: $(5; : 1);(-1; : -5)$