11.(每小题4分,共12分)若 $a=-\frac{1}{2}-\frac{\sqrt{3}}{2},b=-\frac{1}{2}+\frac{\sqrt{3}}{2}$,求下列各式的值:
(1) $a^2-ab+b^2$;
(2) $a^2+b^2$;
(3) $\frac{1}{a}+\frac{1}{b}$.
(1) $a^2-ab+b^2$;
(2) $a^2+b^2$;
(3) $\frac{1}{a}+\frac{1}{b}$.
答案：11.解:$a+b=( -\frac{1}{2}-\frac{\sqrt{3}}{2})+( -\frac{1}{2}+\frac{\sqrt{3}}{2})=-1$,
$ab=( -\frac{1}{2}-\frac{\sqrt{3}}{2})×( -\frac{1}{2}+\frac{\sqrt{3}}{2})=-\frac{1}{2}$.
(1)$a^{2}-ab+b^{2}=(a+b)^{2}-3ab=(-1)^{2}-3×( -\frac{1}{2})=\frac{5}{2}$.
(2)$a^{2}+b^{2}=(a+b)^{2}-2ab=(-1)^{2}-2×( -\frac{1}{2})=1+1=2$.
(3)$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{-1}{-\frac{1}{2}}=2$.
$ab=( -\frac{1}{2}-\frac{\sqrt{3}}{2})×( -\frac{1}{2}+\frac{\sqrt{3}}{2})=-\frac{1}{2}$.
(1)$a^{2}-ab+b^{2}=(a+b)^{2}-3ab=(-1)^{2}-3×( -\frac{1}{2})=\frac{5}{2}$.
(2)$a^{2}+b^{2}=(a+b)^{2}-2ab=(-1)^{2}-2×( -\frac{1}{2})=1+1=2$.
(3)$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{-1}{-\frac{1}{2}}=2$.
12.(10分)如图,某居民小区有一块长方形绿地,长方形绿地的长BC为 $\sqrt{243}$ m,宽AB为 $\sqrt{128}$ m,现要在长方形绿地中修建一个长方形花坛(即图中阴影部分),长方形花坛的长为 $(\sqrt{14}+1)$m,宽为 $(\sqrt{14}-1)$m.
(1)长方形ABCD的周长是多少?
(2)除去修建花坛的地方,其他地方全修建成通道,通道要铺上造价为5元/$m^2$的地砖,铺完整
个通道,购买地砖需要花多少元?(结果化为最简二次根式)
(1)长方形ABCD的周长是多少?
(2)除去修建花坛的地方,其他地方全修建成通道,通道要铺上造价为5元/$m^2$的地砖,铺完整
个通道,购买地砖需要花多少元?(结果化为最简二次根式)
答案：12.解:(1)$2×(\sqrt{243}+\sqrt{128})=2×(9\sqrt{3}+8\sqrt{2})=(18\sqrt{3}+16\sqrt{2})\mathrm{m}$.
答:长方形ABCD的周长是$(18\sqrt{3}+16\sqrt{2})\mathrm{m}$.
(2)$5\left \lbrack \sqrt{243}×\sqrt{128}-(\sqrt{14}+1)(\sqrt{14}-1)\right\rbrack=5×\left \lbrack 72\sqrt{6}-(14-1)\right\rbrack=5×\left \lbrack 72\sqrt{6}-13\right\rbrack=360\sqrt{6}-65$(元).
答:购买地砖需要花费$(360\sqrt{6}-65)$元.
答:长方形ABCD的周长是$(18\sqrt{3}+16\sqrt{2})\mathrm{m}$.
(2)$5\left \lbrack \sqrt{243}×\sqrt{128}-(\sqrt{14}+1)(\sqrt{14}-1)\right\rbrack=5×\left \lbrack 72\sqrt{6}-(14-1)\right\rbrack=5×\left \lbrack 72\sqrt{6}-13\right\rbrack=360\sqrt{6}-65$(元).
答:购买地砖需要花费$(360\sqrt{6}-65)$元.
13.(12分)观察下列各式回答问题:
$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}=1+\frac{1}{1}-\frac{1}{2};\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}=1+\frac{1}{2}-\frac{1}{3};\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}=1+\frac{1}{3}-\frac{1}{4}.$
(1)猜想: $\sqrt{1+\frac{1}{7^2}+\frac{1}{8^2}}=$
(2)归纳:根据你的观察、猜想,写出一个用n(n为正整数)表示的等式: $\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=$
(3)应用:用上述规律计算 $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+\dots+\sqrt{1+\frac{1}{9^2}+\frac{1}{10^2}}$.
$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}=1+\frac{1}{1}-\frac{1}{2};\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}=1+\frac{1}{2}-\frac{1}{3};\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}=1+\frac{1}{3}-\frac{1}{4}.$
(1)猜想: $\sqrt{1+\frac{1}{7^2}+\frac{1}{8^2}}=$
$1+\frac{1}{7}-\frac{1}{8}$
=$1\frac{1}{56}$
;(2)归纳:根据你的观察、猜想,写出一个用n(n为正整数)表示的等式: $\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=$
$1+\frac{1}{n}-\frac{1}{n+1}$
;(3)应用:用上述规律计算 $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+\dots+\sqrt{1+\frac{1}{9^2}+\frac{1}{10^2}}$.
答案：13.(1)$1+\frac{1}{7}-\frac{1}{8}$ $1\frac{1}{56}$ (2)$1+\frac{1}{n}-\frac{1}{n+1}$
(3)解:$\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+\sqrt{1+\frac{1}{3^{2}}+\frac{1}{4^{2}}}+\dots+\sqrt{1+\frac{1}{9^{2}}+\frac{1}{10^{2}}}$
$=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+\dots+1+\frac{1}{9}-\frac{1}{10}$
$=10-\frac{1}{10}=9\frac{9}{10}$.
(3)解:$\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+\sqrt{1+\frac{1}{3^{2}}+\frac{1}{4^{2}}}+\dots+\sqrt{1+\frac{1}{9^{2}}+\frac{1}{10^{2}}}$
$=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+\dots+1+\frac{1}{9}-\frac{1}{10}$
$=10-\frac{1}{10}=9\frac{9}{10}$.