Question

50 за лучший ответ с пояснением. решить две системы.

Answer 1

$\left \{ {{xy- \frac{x}{y} =6} \atop {xy- \frac{y}{x} = \frac{15}{2} }} \right. \\ xy=a \\ \frac{x}{y} =b \\ \left \{ {{a- b=6} \atop {a- \frac{1}{b} = \frac{15}{2} }} \right. \\ \left \{ {{a- b=6} \atop {-a+ \frac{1}{b} = -7.5 }} \right. \\ \frac{1}{b}-b=6-7.5 \\ \frac{1}{b}-b+1.5=0 \\ 1-b^2+1.5b=0 \\ b^2-1.5b-1=0 \\ 2b^2-3b-2=0 \\ D=9+16=25 \\ b= \frac{3+5}{4} =2;a=6+2=8 \\ b= \frac{3-5}{4} =-0.5;a=6-0.5=5.5$
$\left \{ {{xy=8} \atop { \frac{x}{y} =2}} \right. \\ \left \{ {{xy=8} \atop { x=2y}} \right. \\ 2yy=8 \\ y^2=4 \\ \boxed{y=2;x=4} \\ \boxed{y=-2;x=-4}$
$\left \{ {{xy=5.5} \atop { \frac{x}{y} =-0.5}} \right. \\ \left \{ {{xy=5.5} \atop { x=-0.5y}} \right. \\ -0.5yy=5.5 \\ y^2=-11$
квадрат отрицательным не бывает, эта ветка решений не имеет

$\left \{ {{ \frac{ x^{3} }{y} +xy =10} \atop { \frac{ y^{3} }{x} +xy =5/2}} \right. \\ xy=a \\ \frac{x^2}{y^2} =b \\ \left \{ {{ \frac{ x^2xy }{y^2} +xy =10} \atop { \frac{ y^2xy }{x^2} +xy =5/2}} \right. \\ \left \{ {{ ab+a=10} \atop { \frac{ a }{b} +a=5/2}} \right. \\ \left \{ {{ a(b+1)=10} \atop { 2a +2ab=5b}} \right. \\ \left \{ {{ a(b+1)=10} \atop { 2a(1 +b)=5b}} \right. \\ \frac{2a(1 +b)}{a(b+1)} = \frac{5b}{10} \\\ 2= \frac{b}{2} \\ b=4 \\ a= \frac{10}{b+1} = \frac{10}{5}=2$
$\left \{ {{xy=2} \atop { \frac{x^2}{y^2} =4}} \right. \\ \left \{ {{xy=2} \atop { \frac{x}{y} =2}} \right. and \left \{ {{xy=2} \atop { \frac{x}{y} =-2}} \right. \\ \left \{ {{xy=2} \atop { x =2y}} \right. and \left \{ {{xy=2} \atop {x=-2y}} \right. \\\ \left \{ {{2yy=2} \atop { x =2y}} \right. and \left \{ {{-2yy=2} \atop {x=-2y}} \right. \\ \left \{ {{y^2=1} \atop { x =2y}} \right. and \left \{ {{-y^2=1} \atop {x=-2y}} \right.$
вторая система отпадает, так как получился квадрат меньше 0
$\left \{ {{y^2=1} \atop { x =2y}} \right. \\\ \boxed{y=1;x=2*1=2} \\ \boxed{y=-1;x=2*(-1)=-2}$