3 номер срочнооооо, плиссссс

Решите задачу:

$\left \{ {{2^{x+y-1}=1} \atop {3^{xy}=(\frac{1}{3}})^{\sqrt2}} \right. \; \left \{ {{2^{x+y-1}=2^0} \atop {3^{xy}=3^{-\sqrt2}}} \right. \; \left \{ {{x+y-1=0 \atop {xy=-\sqrt2}} \right. \; \left \{ {{y=1-x} \atop {x(1-x)=-\sqrt2}} \right. \\\\x(1-x)=x-x^2\\\\x-x^2=-\sqrt2\\\\x^2-x-\sqrt2=0\\\\D=1+4\sqrt2\\\\x_1= \frac{1-\sqrt{1+4\sqrt2}}{2} \; ,\; \; x_2=\frac{1+\sqrt{1+4\sqrt2}}{2}\\\\y_1=1- \frac{1-\sqrt{1+4\sqrt2}}{2}= \frac{1+\sqrt{1+4\sqrt2}}{2} \; ,$

$y_2=1- \frac{1+\sqrt{1+4\sqrt2}}{2}=\frac{1-\sqrt{1+4\sqrt2}}{2}$

$x_1^2+y_1^2=x_2^2+y_2^2= \frac{(1-\sqrt{1+4\sqrt2})^2}{4} + \frac{(1+\sqrt{1+4\sqrt2})^2}{4} =\\\\= \frac{(1-2\sqrt{1+4\sqrt2}+\, 1+4\sqrt2)+(1+2\sqrt{1+4\sqrt2}+\, 1+4\sqrt2)}{4} =\\\\= \frac{4+8\sqrt2}{4} = \frac{4(1+2\sqrt2)}{4} =1+2\sqrt2$