Question

Determinati a apartine lui R, pt care : x^2 + y^2-x-y+3a > 0

Answer 1

Răspuns:

Explicație pas cu pas:

$x^2+y^2-x-y+3a>0\\\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2-y+\dfrac{1}{4}\right)+3a-\dfrac{1}{2}>0\\\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2+3a-\dfrac{1}{2}>0\\\texttt{Evident , }\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2\geq 0.\texttt{deci conditia este }3a-\dfrac{1}{2}>0,\\\texttt{adica }a>\dfrac{1}{6}.$