Question

(3x+y-4)^2(x+y-2)^2=0

Answer 1

$(3x+y-4)^2+(x+y-2)^2=0$

Так как
$(3x+y-4)^2 geq 0$
и
$(x+y-2)^2 geq 0$
то уравнение верно только при
$left {begin{array}{I} 3x+y-4=0 \ x+y-2=0 end{array}}$

Решаем
$left {begin{array}{I} 3x+y-4=0 \ - \ x+y-2=0 end{array}}$
$2x-2=0 \ x=1 \ \ 3+y-4=0 \ y=1$

Ответ: $(1; 1)$

Answer 2

$(3x+y-4)^2(x+y-2)^2=0 \ left { {{(3x+y-4)^2=0} atop {(x+y-2)^2=0}} right. \ left { {{3x+y-4=0} atop {x+y-2=0}} right. \ -left { {{3x+y-4=0} atop {x+y-2=0}} right. \ 2x-2=0 \ x=1 \ y=2-x; y=2-1=1$
x = 1; y = 1