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Daca | x+y-3 |+ √x^2 - 4x +4 = 0 ,atunci (x;y) = ... ?
Raspunsul este (2;1)


Răspuns :

[tex]|x+y-3|+ \sqrt{(x-2)^2}=0\\ |x+y-3|+ |x-2|=0\\ x+y-3=0\\ x-2=0=\ \textgreater \ x=2\\ 2+y-3=0=\ \textgreater \ y=1\\ (x;y)=(2;1)[/tex]