Odpowiedzi

2009-12-22T23:13:57+01:00
(x+1)^2+(y-3)^2=16 ^2 do potęgi drugiej
x-y=0

x^2+2x+1+y^2-6y+9-16=0
x=y

x^2+2x+1+x^2-6x-7=0
x=y

2x^2-4x-6=0
x=y

delta=b^2-4ac=(-4)^2-4*2*(-6)=64
pierw z delty=8
x1= (-b-pierw z delty)/2a=(4-8)/4= -1=y1
x2= (-b+pierw z delty)/2a=(4+8)/4=3=y2
2009-12-23T00:38:05+01:00
Rozwiaz uklad rownan
{ (x+1)²+(y-3)²=16
x-y=0→→x=y

(x+1)²+(x-3)²=16
x²+2x+1+x²-6x+9=16
2x²-4x-6=0 /:2
x²-2x-3=0
Δ=4+12=16
√Δ=4
x₁=-1, x₂=3
y₁=-1, y₂=3