10. 已知:$x=3+\sqrt{5},y=3-\sqrt{5}$,求下列各式的值.
(1)$x^{2}-y^{2}$;
(2)$\frac{y}{x}+\frac{x}{y}$.
答案：10.解:$\because x=3+\sqrt{5}$,$y=3-\sqrt{5}$,
$\therefore x+y=3+\sqrt{5}+3-\sqrt{5}=6$,
$xy=(3+\sqrt{5})×(3-\sqrt{5})=4$,
$x-y=3+\sqrt{5}-(3-\sqrt{5})=2\sqrt{5}$.
(1)$x^{2}-y^{2}=(x+y)(x-y)=6×2\sqrt{5}=12\sqrt{5}$.
(2)$\frac{y}{x}+\frac{x}{y}=\frac{x^{2}+y^{2}}{xy}=\frac{(x+y)^{2}-2xy}{xy}=\frac{6^{2}-2×4}{4}=$
$\frac{36-8}{4}=7$.
11. (2025·西安期中)如图,张大伯家有一块大长方形空地,长方形空地的长为$\sqrt{98}\ \mathrm{m}$,宽为$\sqrt{32}\ \mathrm{m}$,现要在空地中划出一块长方形地养鸡(即图中阴影部分),其余部分种植蔬菜,长方形养鸡场的长为$(\sqrt{11}+1)\mathrm{m}$,宽为$(\sqrt{11}-1)\mathrm{m}$.
(1)求大长方形空地的周长;(结果化为最简二次根式)
(2)张大伯种植的蔬菜每平方米产量为16千克,求张大伯种植蔬菜的总产量.
答案：11.解:(1)由题意,得大长方形空地的周长为$2(\sqrt{98}+$
$\sqrt{32})=2(7\sqrt{2}+4\sqrt{2})=22\sqrt{2}(\mathrm{m})$.
答:大长方形空地的周长为$22\sqrt{2}\ \mathrm{m}$.
(2)由题意,得种植蔬菜的面积为
$\sqrt{98}×\sqrt{32}-(\sqrt{11}+1)(\sqrt{11}-1)=7\sqrt{2}×4\sqrt{2}-$
$[(\sqrt{11})^{2}-1]=56-10=46(\mathrm{m}^{2})$,
总产量为$16×46=736$(千克).
答:张大伯种植蔬菜的总产量为736千克.
12. 在进行二次根式的化简与运算时,我们有时会碰上如$\frac{3}{\sqrt{5}},\sqrt{\frac{2}{3}},\frac{4}{\sqrt{5}+1}$这样的式子,其实我们还可以将其进一步化简:$\frac{3}{\sqrt{5}}=\frac{3×\sqrt{5}}{\sqrt{5}×\sqrt{5}}=\frac{3\sqrt{5}}{5}$;$\sqrt{\frac{2}{3}}=\sqrt{\frac{2×3}{3×3}}=\frac{\sqrt{6}}{3}$;$\frac{4}{\sqrt{5}+1}=\frac{4(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)}=\sqrt{5}-1$.以上这种化简的步骤叫作分母有理化.
(1)化简:$\frac{1}{\sqrt{3}}=$
$\frac{\sqrt{3}}{3}$
;$\sqrt{\frac{2}{5}}=$
$\frac{\sqrt{10}}{5}$
.
(2)填空:$\sqrt{5}+\sqrt{6}$的倒数为
$\sqrt{6}-\sqrt{5}$
.
(3)化简:$(\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\dots+\frac{1}{\sqrt{2n+1}+\sqrt{2n-1}})×(\sqrt{2n+1}+1)$.
答案：12.(1)$\frac{\sqrt{3}}{3}$ $\frac{\sqrt{10}}{5}$ (2)$\sqrt{6}-\sqrt{5}$
(3)解:原式$=\frac{1}{2}(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+···+$
$\sqrt{2n+1}-\sqrt{2n-1})×(\sqrt{2n+1}+1)=\frac{1}{2}(\sqrt{2n+1}-$
$1)×(\sqrt{2n+1}+1)=\frac{1}{2}(2n+1-1)=n$.
