Решите систему 1)x^3+y^3=152 x^2y+xy^2=120 2)x+y+(x/y)=9 ((x+y)x)/y=20

1)
$\left \{ {x^3+y^3=152} \atop {x^2y+xy^2=120}} \right.$
домножим второе условие системы на 3
$\left \{ {x^3+y^3=152} \atop {3x^2y+3xy^2=360}} \right.$
складываем
$\left \{ {x^3+y^3+3x^2y+3xy^2=512} \atop {x^2y+xy^2=120}} \right.$
$\left \{ {(x+y)^3=8^3} \atop {xy(x+y)=120}} \right.$
$\left \{ {x+y=8} \atop {xy*8=120}} \right.$
$\left \{ {x+y=8} \atop {xy=15}} \right.$
$\left \{ {x=8-y} \atop {y(8-y)=15}} \right.$
$\left \{ {x=8-y} \atop {-y^2+8y-15=0}} \right.$
$\left \{ {x=8-y} \atop {y^2-8y+15=0}} \right.$
$y^2-8y+15=0}}$
$D=(-8)^2-4*1*15=4$
$y_1= \frac{8+2}{2}=5$
$y_2= \frac{8-2}{2}=3$
$\left \{ {{y_1=5} \atop {x_1=8-5}} \right.$ или $\left \{ {{y_2=3} \atop {x_2=8-3}} \right.$
$\left \{ {{y_1=5} \atop {x_1=3}} \right.$ или    $\left \{ {{y_2=3} \atop {x_2=5}} \right.$

Ответ: (3;5) и (5;3)

2)
$\left \{ {{x+y+ \frac{x}{y} =9} \atop { \frac{(x+y)*x}{y} =20}} \right.$
введем замену:
$x+y=a$
$\frac{x}{y} =t$
$\left \{ {{a+t=9} \atop a*t=20}} \right.$
$\left \{ {{a=9-t} \atop (9-t)*t=20}} \right.$
$\left \{ {{a=9-t} \atop t^2-9t+20=0}} \right.$
$t^2-9t+20=0$
$D=(-9)^2-4*1*20=1$
$t_1= \frac{9+1}{2}=5$ , $a_1=9-5=4$
$t_2= \frac{9-1}{2}=4$ , $a_2=9-4=5$

$\left \{ {{x+y=4} \atop { \frac{x}{y} =5}} \right.$ или    $\left \{ {{x+y=5} \atop { \frac{x}{y} =4}} \right.$
$\left \{ {{x+y=4} \atop {x=5y}} \right.$ или    $\left \{ {{x+y=5} \atop {x=4y}} \right.$
$\left \{ {{x=5y} \atop {6y=4}} \right.$    или    $\left \{ {{x=4y} \atop {5y=5}} \right.$
$\left \{ {{y= \frac{2}{3} } \atop {x=3 \frac{1}{3} }} \right.$ или    $\left \{ {{y=1} \atop {x=4}} \right.$

Ответ: $(3 \frac{1}{3}; \frac{2}{3})$ ; $(4;1)$