Wykaż, że jeśli x^{2}+y^{2}=3 i x-y=-2 to xy = 1/2
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(-2)^{2}=(x-y)^{2}= x^{2} + y^{2} -2xy = 3-2xy
=> 4=3-2xy
=> 2xy= (3-4)
=> xy =-1/2
x² + y² = 3
x + y = - 2
x² + y² = (x + y)² - 2xy
3 = (-2)² - 2xy
3 = 4 - 2xy
3 - 4 = - 2xy
-1 = -2xy
1-/(-2) = xy
xy = 1/2 -------------- odpowiedź
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(-2)^{2}=(x-y)^{2}= x^{2} + y^{2} -2xy = 3-2xy
=> 4=3-2xy
=> 2xy= (3-4)
=> xy =-1/2
x² + y² = 3
x + y = - 2
x² + y² = (x + y)² - 2xy
3 = (-2)² - 2xy
3 = 4 - 2xy
3 - 4 = - 2xy
-1 = -2xy
1-/(-2) = xy
xy = 1/2 -------------- odpowiedź