Решите.......................

$\left \{ {{ \sqrt{x} + \sqrt{y} =5} \atop {x+y=13}} \right. \\\ \left \{ {{( \sqrt{x} + \sqrt{y})^2=5^2} \atop {x=13-y}} \right. \\\ x+y+2 \sqrt{xy} =25 \\\ 13+2 \sqrt{xy} =25 \\\ 2 \sqrt{xy} =12 \\\ \sqrt{xy} =6 \\\ xy=36 \\\ (13-y)y=36 \\\ y^2-13y+36=0 \\\ (y-4)(y-9)=0 \\\ y_1=4; \ x_1=13-4=9 \\\ y_2=9; \ x_2=13-9=4$
Ответ: (4; 9); (9; 4)

$\sqrt{x+2} =2+ \sqrt{x-6} \ (x \geq 6) \\\ ( \sqrt{x+2})^2 =(2+ \sqrt{x-6})^2 \\\ x+2=4+x-6+4 \sqrt{x-6} \\\ 4=4 \sqrt{x-6} \\\ 1= \sqrt{x-6} \\\ x-6=1 \\\ x=7$
Ответ: 7