Выполните действия. 4.73 1)

C.4.73.
1)
$(2+\sqrt{7})(\sqrt{7}-2)-(5+\sqrt{2})^2+10\sqrt{2}=\\(\sqrt{7})^2-2^2-(5^2+2*5\sqrt{2}+(\sqrt{2})^2)+10\sqrt{2}=\\7-4-(25+10\sqrt{2}+2)+10\sqrt{2}=3-25-10\sqrt{2}-2+10\sqrt{2}=-24$
2)
$\frac{2}{7}\sqrt{5}*1\frac{3}{4}\sqrt{5}+(\sqrt{25^0}-2)^2=\frac{2}{7}*\frac{7}{4}*(\sqrt{5})^2+(\sqrt{1}-2)^2=\\\frac{5}{2}+(-1)^2=2,5+1=3,5$
3)
$\frac{5}{3-2\sqrt{2}}+\frac{5}{3+2\sqrt{2}}=\frac{5(3+2\sqrt{2})}{(3-2\sqrt{2})(3+2\sqrt{2})}+\frac{5(3-2\sqrt{2})}{(3+2\sqrt{2})(3-2\sqrt{2})}=\\\frac{5(3+2\sqrt{2})+5(3-2\sqrt{2})}{(3-2\sqrt{2})(3+2\sqrt{2})}=\frac{5(3+2\sqrt{2}+3-2\sqrt{2})}{3^2-(2\sqrt{2})^2}=\frac{5*6}{9-8}=30$