Question

daca avem x^2 +y^2-4x+6y+10=-3 atunci (x-y)totul 3

Answer 1

$x^2 +y^2-4x+6y+10=-3 \\ \\ x^2 +y^2-4x+6y+10+3=0 \\ \\ x^2 +y^2-4x+6y+9+4=0\\ \\ x^2 -4x+4+y^2+6y+9=0\\ \\ (x-2)^2+(y+3)^2 = 0 \\ \\ Avem o suma de 2 numere \geq0, singura suma de 2 numere \geq0 care \\ este egala cu 0 este 0+0.\\ \\ \Rightarrow \left\{ \begin{array}{c} (x-2)^2=0 \\ (y+3)^2 = 0 \end{array} \right \Rightarrow \left\{ \begin{array}{c} x-2=0 \\ y+3 = 0 \end{array} \right \Rightarrow \left\{ \begin{array}{c} x=2 \\ y = -3\end{array} \right |$


$\Rightarrow (x-y)^3 = \Big(2-(-3)\Big)^3 =(2+3)^3 = 5^3 = 125$