Question

5x^2+(8y-2)*x+(5y^2+2y+2)=0

Answer 1

Ответ:

$(1; -1)$

Объяснение:

$5x^{2}+8xy-2x+5y^{2}+2y+2=0;$

$5x^{2}+8xy+5y^{2}-2x+2y+2=0;$

$4x^{2}+8xy+4y^{2}+x^{2}+y^{2}-2x+2y+2=0;$

$4(x+y)^{2}+x^{2}+2xy+y^{2}-2x+2-2xy+2y=0;$

$4(x+y)^{2}+(x+y)^{2}-2(x-1)-2y(x-1)=0;$

$5(x+y)^{2}-2(x-1+y(x-1))=0;$

$5(x+y)^{2}=2(y+1)(x-1);$

$(x+y)(x+y)=0,4(y+1)(x-1);$

$\left \{ {{x+y=0,4(y+1)} \atop {x+y=x-1}} \right. \vee \left \{ {{x+y=0,4(x-1)} \atop {x+y=y+1}} \right. \Leftrightarrow \left \{ {{x+y=0,4(y+1)} \atop {y=-1}} \right. \vee \left \{ {{x+y=0,4(x-1)} \atop {x=1}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{x-1=0} \atop {y=-1}} \right. \vee \left \{ {{1+y=0} \atop {x=1}} \right. \Leftrightarrow \left \{ {{x=1} \atop {y=-1}} \right. \vee \left \{ {{y=-1} \atop {x=1}} \right. ;$

$(1; -1);$