Question

(x+1)^2+(y-3)^2 <9
Срочно! Помогите пожалуйста

Answer 1

(x+1)²+(y-3)²<9
(y-3)²+(x+1)²-9<0
x²+2x+1+y²-6y+9-9<0
x²+2x+y²-6y+1<0
x²+2x+y²-6y+1=0
D=√(4-4*(y²-6y+1))=√(4-4y²+24y-4)=√(24y-4y²)=2*√(6y-y²)
x₁=(-2-2√(6y-y²)/2=-(√(6y-y²)-1
x₂=√(6y-y²)-1
(x-√(6y-y²)+1)*(x+√(6y-y²)+1)<0
Так как 6y-y²≥0 ⇒
-∞____+____-√(6y-y²)-1____-____√(6y-y²)-1_____+_____+∞
x∈((-√(6y-y²)-1);((√(6y-y²)-1)).