Question

(х+2у-3)^2+x^2-4xy+4y^2=0
уравнение

Answer 1

$(x+2y-3)^2+x^2-4xy+4y^2=0 \ x^2+4xy+4y^2-6x-12y+9+x^2-4xy+4y^2=0 \ 2x^2+8y^2-6x-12y+9=0 \ 2(x-1.5)^2+8(y-0.75)^2=0 \ left { {{x-1.5=0} atop {y-0.75=0}} right. to left { {{x=1.5} atop {y=0.75}} right.$