Question

ПОМОГИТЕ ПОЖАЛУЙСТА АЛГЕБРА
Решите системы уравнений удобным способом:

Answer 1

1)
$left { {{4x^2+y^2=13} atop {xy=-3}} right.$
$y=- frac{3}{x} \ \ 4x^2+ frac{9}{x^2} =13 \ \ 4x^4-13x^2+9=0 \ \ D=169-4*4*9=25=5^2 \ \ x^2 geq 0 \ \ x^2=(13+5)/8=9/4=(3/2)^2 \ \ x_{1} =3/2 \ y_{1} =-2 \ \ x_{2} =-3/2 \ y_{2} =2 \ \ x^2=(13-5)/8=1 \ \ x_{3} =1 \ y_{3} =-3 \ \ x_{4} =-1 \ y_{4} =3$

2)
$left { {{ frac{x+y}{x-y} - frac{2(x-y)}{x+y} =1} atop {x^2-5xy+2y^2=4}} right.$

$t= frac{x+y}{x-y} \ \ x neq y \ x neq -y \ \ t- frac{2}{t} =1 \ \ t^2-t-2=0 \ \ (t+1)(t-2)=0 \ \ 1)t=-1 \ frac{x+y}{x-y} =-1 \ x+y=-x+y \ 2x=0 \ x=0 \ \ x^2-5xy+2y^2=4 \ 2y^2=4 \ y^2=2 \ y_{1} = sqrt{2} \ y_{2} =- sqrt{2} \ \ 2)t=2 \ x+y=2(x-y) \ x=3y \ \ (3y)^2-5*3y*y+2y^2=4 \ \ 9y^2-15y^2+2y^2=4 \ \ -4y^2=4 \ \ y^2=-1$
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