Question

Решить систему!!!!!!!!!!!

Answer 1

Ответ:

$2$

Пошаговое объяснение:

$\left \{ {{x(3y-x)=2} \atop {9y^2-x^2=5}} \right. \\ \\ \left \{ {{x(3y-x)=2} \atop {(3y)^2-x^2=5}} \right. \\ \\ \left \{ {{x(3y-x)=2} \atop {(3y-x)(3y+x)=5}} \right. \\ \\ \left \{ {{3y-x=\frac{2}{x} } \atop {\frac{2}{x} (3y+x)=5}} \right. \\ \\ \left \{ {{3y-x=\frac{2}{x} } \atop {3y+x=\frac{5x}{2} }} \right. \\ \\ \left \{ {{3y-x=\frac{2}{x} } \atop {3y=\frac{3x}{2} }} \right. \\ \\ \left \{ {{3y-x=\frac{2}{x} } \atop {y=\frac{x}{2} }} \right.$

$\left \{ {{3\cdot \frac{x}{2}-x=\frac{2}{x} } \atop {y=\frac{x}{2} }} \right. \\ \\ \left \{ {{\frac{x}{2}=\frac{2}{x} } \atop {y=\frac{x}{2} }} \right. \\ \\ \left \{ {{x\cdot x=2\cdot 2} \atop {y=\frac{x}{2} }} \right. \\ \\ \left \{ {{x^2=4} \atop {y=\frac{x}{2} }} \right. \\ \\ \left \{ {{x= \pm 2} \atop {y=\frac{x}{2} }} \right.$

$\left \{ {{x=-2} \atop {y=\frac{-2}{2} }} \right.$ или $\left \{ {{x=2} \atop {y=\frac{2}{2} }} \right.$

$\left \{ {{x=-2} \atop {y=-1 }} \right.$ или $\left \{ {{x=2} \atop {y=1 }} \right.$

так как $x>0$, то $x=2,~~y=1$