(0; 1); (2; 1)
Объяснение:
(y(x-1))²+(x-1)²+y²+1-4y|x-1|=0
(y|x-1|)²-2y|x-1|+1+|x-1|²-2y|x-1|+y²=0
(y|x-1|-1)²+(|x-1|-y)²=0
(y|x-1|-1)²≥0, (|x-1|-y)²≥0⇒(y|x-1|-1)²=(|x-1|-y)²=0⇒y|x-1|-1=|x-1|-y=0
y|x-1|-1=0
|x-1|-y=0
y=|x-1|≥0
y·y-1=0
y²=1
y=1
|x-1|=1
x-1=±1
x₁=0, x₂=2
