Question

Решите систему уравнений х^2+у^2=34, х*у=15

Answer 1

x^2+y^2=34

xy=15

_________________

 

y=15/x

225/x^2+x^2=34

225+x^4-34x^2                 t=x^2 t^2=x^4

225+t^2-34t=0 

t^2-34t+225=0

 D=34^2-4*225=256=16^2

t1=34+16/2=25

t2=34-16/2=9

x^2=25=> x1=5 x2=-5

x^2=9=> x3=3 x4=-3

 y1=15/5=3

y2=15/-5=-3

y3=15/3=5

y4=15/-5=-3

Otvet: x1=5 y1=3 x2=-5 y2=-3 x3=3 y3=5 x4=-3 y4=-5 

 

 

 

 

Answer 2

$ttdisplaystyle left { {{x^2+y^2=34} atop {xy=15}} right. Longrightarrow left { {{x^2+y^2=34} atop {y=frac{15}{x}} right.$

Подставляем.

$ttdisplaystyle x^2+(frac{15}{x})^2=34\\\x^2+frac{225}{x^2}=34|*x^2\\\x^4+225-34x^2=0\\x^2=t\\t^2-34t+225=0\\D=(-34)^2-4*1*225=1156-900=256\\sqrt{256}=16\\t_1=frac{34-16}{2}=9\\\t_2=frac{34+16}{2}=25$

x² = t

x² = 9 или x² = 25

x = ±3        x² = ±5

y₁ = 15/3 = 5

y₂ = 15/(-3) = -5

y₃ = 15/5 = 3

y₄ = 15/(-5) = -3

Ответ: (3; 5); (-3; -5); (5; 3); (-5; -3)