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cat1111
@cat1111
July 2022
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Решите систему уравнений
х в квадрате+у в квадрате=10
ху=3
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prostotom
X²+y²=10
xy=3 ⇒ y=3/x
x²+(3/x)²=10,
x²+9/x²-10=0,
x⁴-10x²+9=0, пусть x²=t, t≥0, тогда
t²-10t+9=0, по теореме Виета t₁=1, t₂=9.
1) x²=1 ⇒ x₁=-1, x₂=1; y₁=3/(-1)=-3, y₂=3/1=3.
2) x²=9 ⇒ x₃=-3, x₄=3; y₃=3/(-3)=-1, y₄=3/3=1.
Ответ: (-1;-3), (1;3), (-3;-1), (3;1).
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Answers & Comments
xy=3 ⇒ y=3/x
x²+(3/x)²=10,
x²+9/x²-10=0,
x⁴-10x²+9=0, пусть x²=t, t≥0, тогда
t²-10t+9=0, по теореме Виета t₁=1, t₂=9.
1) x²=1 ⇒ x₁=-1, x₂=1; y₁=3/(-1)=-3, y₂=3/1=3.
2) x²=9 ⇒ x₃=-3, x₄=3; y₃=3/(-3)=-1, y₄=3/3=1.
Ответ: (-1;-3), (1;3), (-3;-1), (3;1).