Question

Решите систему уравнений (х+у)в квадрате=2у
(х+у)в квадрате=2х

Answer 1

Объяснение:

$\left \{ {(x+y)^2=2y} \atop {x+y=2x}} \right.$$\left \{(x+y)^2=2y} \atop {y=2x-x}} \right.$$\left \{ {{(x+y)^2=2y} \atop {y=x}} \right.$⇒$(y+y)^{2}=2y\\(2y)^2=2y\\4y^2=2y \\2y^2=y\\2y^2-y=0\\y(2y-1)=0\\y=0; 2y-1=0\\y=0; y=\frac{1}{2}$

$0=x\\x=\frac{1}{2}$⇒$x=0\\x=\frac{1}{2}$

$\left \{ {{(0+0)^2=2*0} \atop {0=0}} \right.$$\left \{ {{0=0} \atop {0=0}} \right.$

$\left \{ {{(\frac{1}{2}+\frac{1}{2})^2 =2*\frac{1}{2} } \atop {\frac{1}{2}=\frac{1}{2} }} \right.$⇒$\left \{ {{1=1} \atop {\frac{1}{2}=\frac{1}{2} }} \right.$

(x1,y2)=(0,0)

(x2,y2)=$(\frac{1}{2} ,\frac{1}{2})$