Question

ребят помогите пожалуйста.
прошу подробное решение

Answer 1

$left { {{4x^{2}+3xy-y^{2}=0} atop {x^{2}-2xy+y^{2}=16}} right.$

$left { {{4x^{2}+3xy-y^{2}=0} atop {(x-y)^{2}=16}} right.$

$left { {{4x^{2}+3xy-y^{2}=0} atop {x-y=4, x-y=-4}} right.$

$left { {{4x^{2}+3xy-y^{2}=0} atop {x=4-y, x=y-4}} right.$

при х=4-y
$left { {{4*(4-y)^{2}+3y*(4-y)-y^{2}=0} atop {x=4-y}} right.$

$left { {{4*(16-8y+y^{2})+12y-3y^{2}-y^{2}=0} atop {x=4-y}} right.$

$left { {{64-32y+4y^{2}+12y-4y^{2}=0} atop {x=4-y}} right.$

$left { {{64-20y=0} atop {x=4-y}} right.$

$left { {{y=frac{16}{5}} atop {x=4-frac{16}{5}=frac{4}{5}}} right.$

при х=y-4
$left { {{4*(y-4)^{2}+3y*(y-4)-y^{2}=0} atop {x=y-4}} right.$

$left { {{4y^{2}-32y+64+3y^{2}-12y-y^{2}=0} atop {x=y-4}} right.$

$left { {{6y^{2}-44y+64=0} atop {x=y-4}} right.$

$left { {{3y^{2}-22y+32=0, D=100>0} atop {x=y-4}} right.$

$left { {{y_{1}= frac{22-10}{6}=2} atop {x=2-4=-2}} right.$

$left { {{y_{2}=frac{16}{3}} atop {x=frac{16}{3}-4=frac{4}{3}}} right.$

Ответ: (4/5; 16/5); (-2; 2); (4/3; 16/3)