Question

Помогите решить, пожалуйста ​

Answer 1

Ответ:

$(2;1);(3;1);(-2;-1);(-3;-1);(3\sqrt{2};\sqrt{2});(2\sqrt{2};\sqrt{2});(-2\sqrt{2};-\sqrt{2});(-3\sqrt{2};-\sqrt{2});$

Пошаговое объяснение:

$4) \left \{ {{\frac{x}{y}+\frac{6y}{x}=5} \atop {x^{2}+4xy-3y^{2}=18}} \right.$

$\left \{ {{\frac{x}{y}*xy+\frac{6y}{x}*xy=5*xy} \atop {x^{2}+4xy-3y^{2}=18}} \right.$

$\left \{ {{x^{2}-5xy+6y^{2}=0} \atop {x^{2}+4xy-3y^{2}=18}} \right.$

$x^{2}-5xy+6y^{2}=0;$

$a=1, b=-5y, c=6y^{2};$

$D=b^{2}-4*a*c;$

$D=(-5y)^{2}-4*1*6y^{2}=25y^{2}-24y^{2}=y^{2};$

$x_{1}=\frac{-b+\sqrt{D}}{2a};$

$x_{1}=\frac{-(-5y)+\sqrt{y^{2}}}{2*1}=\frac{5y+y}{2}=\frac{6y}{2}=3y;$

$x_{2}=\frac{-b-\sqrt{D}}{2a};$

$x_{2}=\frac{-(-5y)-\sqrt{y^{2}}}{2*1}=\frac{5y-y}{2}=\frac{4y}{2}=2y;$

$\left \{ {{x^{2}-5xy+6y^{2}=0} \atop {(3y)^{2}+4*3y*y-3y^{2}=18}} \right.$

$\left \{ {{x^{2}-5xy+6y^{2}=0} \atop {9y^{2}+12y^{2}-3y^{2}=18}} \right.$

$\left \{ {{x^{2}-5xy+6y^{2}=0} \atop {18y^{2}=18}} \right.$

$\left \{ {{x^{2}-5xy+6y^{2}=0} \atop {y^{2}=1}} \right.$

$\left \{ {{x^{2}-5xy+6y^{2}=0} \atop {y=1}} \right. ; \left \{ {{x^{2}-5xy+6y^{2}=0} \atop {y=-1}} \right.$

$\left \{ {{x^{2}-5x+6=0} \atop {y=1}} \right. ; \left \{ {{x^{2}+5x+6=0} \atop {y=-1}} \right.$

$\left \{ {{(x-2)(x-3)=0} \atop {y=1}} \right. ; \left \{ {{(x+2)(x+3)=0} \atop {y=-1}} \right.$

$\left \{ {{x=2} \atop {y=1}} \right. ; \left \{ {{x=3} \atop {y=1}} \right. ; \left \{ {{x=-2} \atop {y=-1}} \right. ; \left \{ {{x=-3} \atop {y=-1}} \right.$

$(2;1);(3;1);(-2;-1);(-3;-1);$

$\left \{ {{x^{2}-5xy+6y^{2}=0} \atop {(2y)^{2}+4*2y*y-3y^{2}=18}} \right.$

$\left \{ {{x^{2}-5xy+6y^{2}=0} \atop {4y^{2}+8y^{2}-3y^{2}=18}} \right.$

$\left \{ {{x^{2}-5xy+6y^{2}=0} \atop {9y^{2}=18}} \right.$

$\left \{ {{x^{2}-5xy+6y^{2}=0} \atop {y^{2}=2}} \right.$

$\left \{ {{x^{2}-5xy+6y^{2}=0} \atop {y=\sqrt{2}}} \right. ; \left \{ {{x^{2}-5xy+6y^{2}=0} \atop {y=-\sqrt{2}}} \right.$

$\left \{ {{x^{2}-5x\sqrt{2}+12=0} \atop {y=\sqrt{2}}} \right. ; \left \{ {{x^{2}+5x\sqrt{2}+12=0} \atop {y=-\sqrt{2}}} \right.$

$x^{2}-5x\sqrt{2}+12=0;$

$D=b^{2}-4*a*c;$

$D=(-5\sqrt{2})^{2}-4*1*12=(-1)^{2}*5^{2}*(\sqrt{2})^{2}-48=1*25*2-48=50-48=2;$

$x_{1}=\frac{-b+\sqrt{D}}{2a};$

$x_{1}=\frac{-(-5\sqrt{2})+\sqrt{2}}{2*1}=\frac{5\sqrt{2}+\sqrt{2}}{2}=\frac{6\sqrt{2}}{2}=3\sqrt{2};$

$x_{2}=\frac{-b-\sqrt{D}}{2a};$

$x_{2}=\frac{-(-5\sqrt{2})-\sqrt{2}}{2*1}=\frac{5\sqrt{2}-\sqrt{2}}{2}=\frac{4\sqrt{2}}{2}=2\sqrt{2};$

$\left \{ {{x=3\sqrt{2}} \atop {y=\sqrt{2}}} \right. ; \left \{ {{x=2\sqrt{2}} \atop {y=\sqrt{2}}} \right.$

$x^{2}+5x\sqrt{2}+12=0;$

$D=b^{2}-4*a*c;$

$D=(5\sqrt{2})^{2}-4*1*12=5^{2}*(\sqrt{2})^{2}-48=25*2-48=50-48=2;$

$x_{1}=\frac{-5\sqrt{2}+\sqrt{2}}{2*1}=\frac{-4\sqrt{2}}{2}=-2\sqrt{2};$

$x_{2}=\frac{-5\sqrt{2}-\sqrt{2}}{2*1}=\frac{-6\sqrt{2}}{2}=-3\sqrt{2};$

$\left \{ {{x=-2\sqrt{2}} \atop {y=-\sqrt{2}}} \right. ; \left \{ {{x=-3\sqrt{2}} \atop {y=-\sqrt{2}}} \right.$

$(2;1);(3;1);(-2;-1);(-3;-1);(3\sqrt{2};\sqrt{2});(2\sqrt{2};\sqrt{2});(-2\sqrt{2};-\sqrt{2});(-3\sqrt{2};-\sqrt{2});$