Question

Daca tripletul (x,y,z) este solutie a sistemului x+y+x=2 si xy+xz+yz=0
atunci aflati valoarea expresiei x^2+y^2+z^2

Answer 1

$(x+y+z)^2 = \Big((x+y)+z\Big)^2 = (x+y)^2+2(x+y)z+z^2 = \\ \\ = x^2+2xy+y^2+2xz+2yz+z^2 = x^2+y^2+z^2+2(xy+xz+yz) \\ \\\\ \Rightarrow (x+y+z)^2 = x^2+y^2+z^2+2(xy+xz+yz) \Rightarrow \\ \\ \Rightarrow x^2+y^2+z^2 = (x+y+z)^2-2(xy+xz+yz) \Rightarrow \\ \\ \Rightarrow x^2+y^2+z^2 = 2^2-2\cdot 0 \Rightarrow \boxed{x^2+y^2+z^2 = 4}$

Answer 2

(x+y+z)²=x²+y²+z²+2xy+2yz+2xz=4 ⇔x²+y²+z²=4 ;