Question

Нужно решить ситему уравнений 1.[x/y+1=6y/x [x-y=3 2.[1/x-1/y=1 [x-11xy-y=-1

Answer 1

$\left \{ {{ \frac{x}{y} +1= \frac{6y}{x} } \atop {x-y=3}} \right.$
$\left \{ {{ \frac{x}{y} +1= \frac{6y}{x} } \atop {x=3+y}} \right.$
$\left \{ {{ \frac{3+y}{y} +1= \frac{6y}{3+y} } \atop {x=3+y}} \right.$
$\left \{ {{ \frac{(3+y)(3+y)+y(3+y)-6y*y}{y(3+y)} =0 } \atop {x=3+y}} \right.$
$\left \{ {{ \frac{9+6y+ y^{2} +3y+ y^{2} -6 y^{2} }{y(3+y)} =0 } \atop {x=3+y}} \right.$
$\left \{ {{ \frac{-4 y^{2}+ 9y+9}{y(3+y)} =0 } \atop {x=3+y}} \right.$
$\left \{ {{ \frac{4 y^{2}- 9y-9}{y(3+y)} =0 } \atop {x=3+y}} \right.$
$4 y^{2}- 9y-9=0$   ОДЗ:$y\neq 0$;  y≠-3
D=81+4*4*9=81+144=225
$y_{1}= \frac{9+15}{8}=3$
$y_{2}= \frac{9-15}{8}=- \frac{3}{4}$
$\left \{ {{y=3} \atop {x=6}} \right.$
$\left \{ {{y=- \frac{3}{4} } \atop {x=2,25 }} \right.$
ответ: (6;3), (2,25; -0,75)

$\left \{ {{ \frac{1}{x} - \frac{1}{y} =1} \atop {x-11xy-y=-1}} \right.$
$\left \{ {{ \frac{y-x-xy}{xy}=0} \atop {x-11xy-y=-1}} \right.$
ОДЗ: х≠0, у≠0
$\left \{ {y-x-xy=0} \atop {x-11xy-y=-1}} \right.$
$\left \{ {-x-xy+y=0} \atop {x-11xy-y=-1}} \right.$
Сложим первое и второе уравнения системы: 
$-12xy=-1$
$x= \frac{1}{12y}$
$\left \{ {-x-xy+y=0} \atop {x= \frac{1}{12y} }} \right.$
$\left \{ {- \frac{1}{12y} - \frac{1}{12} +y=0} \atop {x= \frac{1}{12y} }} \right.$
$\left \{ { \frac{-1-y+12 y^{2} }{12y}=0} \atop {x= \frac{1}{12y} }} \right.$
$-1-y+12 y^{2} =0$
ОДЗ: у≠0
D=1+4*12*1=49
$y_{1} = \frac{1+7}{24} = \frac{1}{3}$
$y_{2} = \frac{1-7}{24} =- \frac{1}{4}$
$\left \{ {{y= \frac{1}{3} } \atop {x= \frac{1}{4} }} \right.$
$\left \{ {{y=- \frac{1}{4} } \atop {x= -\frac{1}{3} }} \right.$
ответ: $( \frac{1}{4}; \frac{1}{3} )$, $(- \frac{1}{3}; -\frac{1}{4} )$