• Предмет: Алгебра
  • Автор: Аноним
  • Вопрос задан 8 лет назад

решите системы уравнений
{x+y=1
x^2+y^2=25

{x+y=3
x^2+y^2=29

Ответы

Ответ дал: NNNLLL54
0
1)quad  left { {{x+y=1} atop {x^2+y^2=25}} right. ;  left { {{(x+y)^2=1} atop {x^2+y^2=25}} right. ;  left { {{x^2+y^2+2xy=1} atop {x^2+y^2=25}} right. ;  left { {{25+2xy=1} atop {x^2+y^2=25}} right. \\ left { {{2xy=-24} atop {x^2+y^2=25}} right. ;  left { {{xy=-12} atop {x^2+y^2=25}} right. ;   left { {{y=frac{-12}{x}} atop {x^2+frac{144}{x^2}=25}} right. ;  left { {{y=-frac{12}{x}} atop {x^4-25x^2+144=0; ,; xne 0}} right.

 left { {{y=-frac{12}{x}} atop {x^2=9; ,; x^2=16}} right. ;  left { {{y_{1,2}=mp 4; ,; x_{3,4}=mp 3} atop {x_{1,2}=pm 3; ,; x_{3,4}=pm 4}} right. \\Otvet:; (3,-4); ,; (-3,4); ,; (4,-3); ,; (-4,3); .

2)quad  left { {{x+y=3} atop {x^2+y^2=29}} right. ;  left { {{(x+y)^2=9} atop {x^2+y^2=29}} right. ;  left { {{9=29+2xy} atop {x^2+y^2=29}} right. ;  left { {{xy=-10} atop {x^2+y^2=29}} right. \\ left { {{y=-frac{10}{x}} atop {x^2+frac{100}{x^2}=29}} right. ;  left { {{y=-frac{10}{x}} atop {x^4-29x^2+100=0,; xne 0}} right. ;  left { {{y=-frac{10}{x}} atop {x^2=4; ,; x^2=25}} right.

   left { {{y=-frac{10}{x}} atop {x_{1,2}=pm 2; ,; x_{3,4}=pm 5} right. \\   left { {{y_{1,2}=mp 5; ,; y_{3,4}=mp 2} atop {x_{1,2}=pm 2; ,; x_{3,4}=pm 5}} right.  

Otvet:; ; (2,-5); ,; (-2,5); ,; (5,-2); ,; (-5,2); .
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