ПОМОГИТЕ ПОЖАЛУЙСТА АЛГЕБРА Решите системы уравнений удобным способом:

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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