[tex]x+y = \sqrt3 \quad $si$ \quad x-y = 1 \\ \\ x^4-y^4 = {(x^2)}^2 - {(y^2)}^2 = (x^2-y^2)(x^2+y^2) = \\ =\Big[(x-y)(x+y)\Big](x^2+y^2) = (x-y)(x+y)(x^2+y^2) = \\ = 1\cdot \sqrt3\cdot(x^2+y^2) = \sqrt3\cdot(x^2+y^2) \\ \\ \left\{ \begin{array}{c} (x+y)^2 =x^2+2xy+y^2 \\ (x-y)^2 = x^2-2xy+y^2 \end{array} \right|_{\big{$adunam$}} \Rightarrow \\ \\ \Rightarrow (x+y)^2+(x-y)^2 = 2x^2+2y^2 \Rightarrow \sqrt3^2+1^2 = 2(x^2+y^2) \Rightarrow \\ [/tex]
[tex]\Rightarrow 3+1 = 2(x^2+y^2) \Rightarrow 4 = 2(x^2+y^2) \Rightarrow x^2+y^2 = \dfrac{4}{2} \Rightarrow \\ \Rightarrow x^2+y^2 = 2 \\ \\ \\ x^4-y^4 = \sqrt3\cdot(x^2+y^2) \Rightarrow x^4-y^4 = \sqrt3\cdot 2 \Rightarrow \boxed{x^4-y^4 = 2\sqrt3} [/tex]
