(x²+3y²-7)²+√(3-xy-y²)=0
a=(x²+3y²-7)²≥0
b=√(3-xy-y²)≥0 a+b=0⇔ a=0 и b=0
⇔(x²+3y²-7)=0 и
(3-xy-y²)=0
x=(3-y²)/y
[(3-y²)/y]²+3y²-7=0 ⇔ 4y⁴-13y²+9=0 1)y²=1 y1=-1 y2=1
x1=-2 x2=2
2)y²=9/4 y3=-3/2 y4=3/2
x3=-1/2 x4=1/2
ппроверка
(-2;-1)
(x²+3y²-7)=0 и 4+3-7=0
(3-xy-y²)=0 3-2-1=0
(2;1)
(x²+3y²-7)=0 и 4+3-7=0
(3-xy-y²)=0 3-2-1=0
(-1/2;-3/2)
(x²+3y²-7)=0 и 1/4+3·(9/4)-7=0
(3-xy-y²)=0 3-3/4-9/4=0
(1/2;3/2)
(x²+3y²-7)=0 и 1/4+3·(9/4)-7=0
(3-xy-y²)=0 3-3/4-9/4=0