Question

решите систему уравнений x-y=4;x^2+y^2=36-2xy

Answer 1

$begin{cases}x-y=4\x^2+y^2=36-2xyend{cases}<=>begin{cases}x-y=4\x^2+y^2+2xy=36end{cases}<=>\<=>begin{cases}x-y=4=>x=4+y\(x+y)^2=36end{cases}\(4+y+y)^2=36\(4+2y)^2=36\4+2y=6 ;4+2y=-6\2y=2 ;2y=-10\y=1 ;y=-5\x=5 x=-1\OTBET:(5;1);(-1;-5)$