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July 2022 1 3 Report

Решить систему уравнений с двумя переменными:

[tex] \left \{ {{ x^{2}y+y^{2}x=20} \atop {x^{3}+ y^{3} =65}} \right. [/tex]

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mukus13

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$\left \{ {{ x^{2} y+ y^{2}x =20}|*3\atop {x^3+y^3=65}} \right.$

$\left \{ {{ 3x^{2} y+ 3y^{2}x =60} \atop {x^3+y^3=65}}$

Складываем (1) и (2) условие системы:

$\left \{ {{3 x^{2} y+3 y^{2}x+x^3+y^3 =60+65} \atop {x^2y+y^2x=20}} \right.$

$\left \{ {{(x+y)^3=125} \atop {xy(x+y)=20}} \right.$

$\left \{ {{(x+y)^3=5^3} \atop {xy(x+y)=20}} \right.$

$\left \{ {{x+y=5} \atop {xy*5=20}} \right.$

$\left \{ {{x+y=5} \atop {xy=4}} \right.$

$\left \{ {{x=5-y} \atop {(5-y)*y=4}} \right.$

$\left \{ {{x=5-y} \atop {-y^2+5y-4=0}} \right.$

$\left \{ {{x=5-y} \atop {y^2-5y+4=0}} \right.$

$D=(-5)^2-4*1*4=25-16=9$
$y_1= \frac{5+3}{2} =4$
$y_2= \frac{5-3}{2} =1$

$x_1=5-4=1$
$x_2=5-1=4$

Ответ: $(1;4)$ ; $(4;1)$

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