Решите систему 1)x^3+y^3=152 x^2y+xy^2=120 2)x+y+(x/y)=9 ((x+y)x)/y=20
Решите систему
1)x^3+y^3=152
x^2y+xy^2=120
2)x+y+(x/y)=9
((x+y)x)/y=20
Ответ(ы) на вопрос:
Гость
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)
Гость
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]
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