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 {(x+y)(x²-xy+y²)=152⇒x+y=152/(x²-xy+y2) {xy(x+y)=120⇒x+y=120/xy 152/(x²-xy+y²)=120/xy 152xy=120x²-120xy+120y² 120x²-272xy+120y²=0 15x²-34xy+15y²=0 решим относительно х D=1156y²-900y²=256y² x1=(34y-16y)/30=3y/5 U x2=(34y+16y)/30=5y/3 1)x=3y/5 27y³/125+y³=152 27y³+125y³=152*125 152y³=152*125 y³=125 y1=5 x1=3/5*5=3 2)x=5y/3 125y³/27+y³=152 152y³=152*27 y³=27 y2=3 x2=5/3*3=5 Ответ (3;5);(5;3) 2 {x+y+x/y=9 {(x+y)*x/y=20 x+y=a  x/y=b {a+b=9 {a*b=20 применим теорему Виета 1)a=4, b=5   U  2)a=5 ,b=4 1){x+y=4 {x/y=5⇒x=5y 5y+y=4 6y=4 y1=2/3 x1=5*2/3=10/3 2){x+y=5 {x/y=4⇒x=4y 4y+y=5 5y=5 y2=1 x2=4*1=4 Ответ (10/3;2/3);(4;1)   

1) [latex] \left \{ {x^3+y^3=152} \atop {x^2y+xy^2=120}} \right. [/latex] домножим второе условие системы на 3 [latex] \left \{ {x^3+y^3=152} \atop {3x^2y+3xy^2=360}} \right. [/latex] складываем [latex] \left \{ {x^3+y^3+3x^2y+3xy^2=512} \atop {x^2y+xy^2=120}} \right. [/latex] [latex] \left \{ {(x+y)^3=8^3} \atop {xy(x+y)=120}} \right. [/latex] [latex] \left \{ {x+y=8} \atop {xy*8=120}} \right. [/latex] [latex] \left \{ {x+y=8} \atop {xy=15}} \right. [/latex] [latex] \left \{ {x=8-y} \atop {y(8-y)=15}} \right. [/latex] [latex] \left \{ {x=8-y} \atop {-y^2+8y-15=0}} \right. [/latex] [latex] \left \{ {x=8-y} \atop {y^2-8y+15=0}} \right. [/latex] [latex]y^2-8y+15=0}} [/latex] [latex]D=(-8)^2-4*1*15=4[/latex] [latex]y_1= \frac{8+2}{2}=5 [/latex] [latex]y_2= \frac{8-2}{2}=3 [/latex] [latex] \left \{ {{y_1=5} \atop {x_1=8-5}} \right. [/latex]   или    [latex] \left \{ {{y_2=3} \atop {x_2=8-3}} \right. [/latex] [latex] \left \{ {{y_1=5} \atop {x_1=3}} \right. [/latex]      или     [latex] \left \{ {{y_2=3} \atop {x_2=5}} \right. [/latex] Ответ: (3;5) и (5;3) 2) [latex] \left \{ {{x+y+ \frac{x}{y} =9} \atop { \frac{(x+y)*x}{y} =20}} \right. [/latex] введем замену:  [latex]x+y=a[/latex] [latex] \frac{x}{y} =t[/latex] [latex] \left \{ {{a+t=9} \atop a*t=20}} \right. [/latex] [latex] \left \{ {{a=9-t} \atop (9-t)*t=20}} \right. [/latex] [latex] \left \{ {{a=9-t} \atop t^2-9t+20=0}} \right. [/latex] [latex]t^2-9t+20=0[/latex] [latex]D=(-9)^2-4*1*20=1[/latex] [latex]t_1= \frac{9+1}{2}=5 [/latex],    [latex]a_1=9-5=4[/latex] [latex]t_2= \frac{9-1}{2}=4 [/latex],     [latex]a_2=9-4=5[/latex] [latex]\left \{ {{x+y=4} \atop { \frac{x}{y} =5}} \right. [/latex]     или     [latex]\left \{ {{x+y=5} \atop { \frac{x}{y} =4}} \right. [/latex] [latex]\left \{ {{x+y=4} \atop {x=5y}} \right. [/latex]      или   [latex]\left \{ {{x+y=5} \atop {x=4y}} \right. [/latex] [latex] \left \{ {{x=5y} \atop {6y=4}} \right. [/latex]        или   [latex]\left \{ {{x=4y} \atop {5y=5}} \right. [/latex] [latex]\left \{ {{y= \frac{2}{3} } \atop {x=3 \frac{1}{3} }} \right. [/latex]      или     [latex]\left \{ {{y=1} \atop {x=4}} \right. [/latex] Ответ: [latex](3 \frac{1}{3}; \frac{2}{3}) [/latex] ; [latex](4;1)[/latex]