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7. 已知$a=\sqrt {3+\sqrt {5}},b=\sqrt {3-\sqrt {5}}$,则$a-b=$(
A. $\sqrt {2}$
B. $\sqrt {3}$
C. $\sqrt {5}$
D. $2\sqrt {2}$
$\sqrt{2}$
)A. $\sqrt {2}$
B. $\sqrt {3}$
C. $\sqrt {5}$
D. $2\sqrt {2}$
答案：
A 解析:$\because a=\sqrt {3+\sqrt {5}},b=\sqrt {3-\sqrt {5}}$,
$\therefore a^{2}=(\sqrt {3+\sqrt {5}})^{2}=3+\sqrt {5},b^{2}=(\sqrt {3-\sqrt {5}})^{2}=3-\sqrt {5},ab=\sqrt {3+\sqrt {5}}\cdot \sqrt {3-\sqrt {5}}=\sqrt {(3+\sqrt {5})(3-\sqrt {5})}=\sqrt {4}=2$.
$\because (a-b)^{2}=a^{2}+b^{2}-2ab=3+\sqrt {5}+3-\sqrt {5}-2×2=2,\therefore a-b=\pm \sqrt {2}$,
$\because 3+\sqrt {5}＞3-\sqrt {5}$,
$\therefore \sqrt {3+\sqrt {5}}＞\sqrt {3-\sqrt {5}}$,
$\therefore a＞b,\therefore a-b＞0,\therefore a-b=\sqrt {2}$.
$\therefore a^{2}=(\sqrt {3+\sqrt {5}})^{2}=3+\sqrt {5},b^{2}=(\sqrt {3-\sqrt {5}})^{2}=3-\sqrt {5},ab=\sqrt {3+\sqrt {5}}\cdot \sqrt {3-\sqrt {5}}=\sqrt {(3+\sqrt {5})(3-\sqrt {5})}=\sqrt {4}=2$.
$\because (a-b)^{2}=a^{2}+b^{2}-2ab=3+\sqrt {5}+3-\sqrt {5}-2×2=2,\therefore a-b=\pm \sqrt {2}$,
$\because 3+\sqrt {5}＞3-\sqrt {5}$,
$\therefore \sqrt {3+\sqrt {5}}＞\sqrt {3-\sqrt {5}}$,
$\therefore a＞b,\therefore a-b＞0,\therefore a-b=\sqrt {2}$.
8. 若$a+\frac {1}{a}=11(0＜a＜1)$,则$\sqrt {a}-\frac {1}{\sqrt {a}}=$
-3
.
答案：
-3 解析:$\because (\sqrt {a}-\frac {1}{\sqrt {a}})^{2}=a+\frac {1}{a}-2=11-2=9,\therefore \sqrt {a}-\frac {1}{\sqrt {a}}=\pm 3$.
又$0＜a＜1,\therefore \sqrt {a}-\frac {1}{\sqrt {a}}＜0.\therefore \sqrt {a}-\frac {1}{\sqrt {a}}=-3$.
又$0＜a＜1,\therefore \sqrt {a}-\frac {1}{\sqrt {a}}＜0.\therefore \sqrt {a}-\frac {1}{\sqrt {a}}=-3$.
9. 解方程:$\frac {2}{3}\sqrt {9x}+6\sqrt {\frac {x}{4}}-3x\sqrt {\frac {1}{x}}=14$.
答案：
$x=49$
10. 已知最简二次根式$\sqrt {2a-4}$与$\sqrt {b}$的被开方数相同,求$\sqrt {4a^{2}-4ab+b^{2}}$的值.
答案：
解:由题意可知$2a-4=b$,
即$2a-b=4$,
则$\sqrt {4a^{2}-4ab+b^{2}}=\sqrt {(2a-b)^{2}}=4$.
即$2a-b=4$,
则$\sqrt {4a^{2}-4ab+b^{2}}=\sqrt {(2a-b)^{2}}=4$.
11.【阅读理解】
爱思考的小明遇到一个问题:已知$a=\frac {1}{2+\sqrt {3}}$,求$2a^{2}-8a+1$的值.
他是这样分析与解答的.
$\because a=\frac {1}{2+\sqrt {3}}=\frac {2-\sqrt {3}}{(2+\sqrt {3})(2-\sqrt {3})}=2-\sqrt {3},\therefore a-2=-\sqrt {3}$.
$\therefore (a-2)^{2}=3$,即$a^{2}-4a+4=3$.
$\therefore a^{2}-4a=-1$.
$\therefore 2a^{2}-8a+1=2(a^{2}-4a)+1=2×(-1)+1=-1$.
请你根据小明的分析过程,解决如下问题.
(1)计算:$\frac {1}{\sqrt {2}+1}=$
(2)计算:$\frac {1}{\sqrt {2}+1}+\frac {1}{\sqrt {3}+\sqrt {2}}+\frac {1}{\sqrt {4}+\sqrt {3}}+... +\frac {1}{\sqrt {100}+\sqrt {99}}=$
(3)若$a=\frac {1}{\sqrt {5}-2}$,求$3a^{2}-12a-1$的值.
爱思考的小明遇到一个问题:已知$a=\frac {1}{2+\sqrt {3}}$,求$2a^{2}-8a+1$的值.
他是这样分析与解答的.
$\because a=\frac {1}{2+\sqrt {3}}=\frac {2-\sqrt {3}}{(2+\sqrt {3})(2-\sqrt {3})}=2-\sqrt {3},\therefore a-2=-\sqrt {3}$.
$\therefore (a-2)^{2}=3$,即$a^{2}-4a+4=3$.
$\therefore a^{2}-4a=-1$.
$\therefore 2a^{2}-8a+1=2(a^{2}-4a)+1=2×(-1)+1=-1$.
请你根据小明的分析过程,解决如下问题.
(1)计算:$\frac {1}{\sqrt {2}+1}=$
$\sqrt{2}-1$
;(2)计算:$\frac {1}{\sqrt {2}+1}+\frac {1}{\sqrt {3}+\sqrt {2}}+\frac {1}{\sqrt {4}+\sqrt {3}}+... +\frac {1}{\sqrt {100}+\sqrt {99}}=$
9
;(3)若$a=\frac {1}{\sqrt {5}-2}$,求$3a^{2}-12a-1$的值.
2
答案：
解:
(1)$\frac {1}{\sqrt {2}+1}=\frac {\sqrt {2}-1}{(\sqrt {2}+1)(\sqrt {2}-1)}=\sqrt {2}-1$.
(2)原式$=\frac {\sqrt {2}-1}{(\sqrt {2}+1)(\sqrt {2}-1)}+\frac {\sqrt {3}-\sqrt {2}}{(\sqrt {3}+\sqrt {2})(\sqrt {3}-\sqrt {2})}+\frac {\sqrt {4}-\sqrt {3}}{(\sqrt {4}+\sqrt {3})(\sqrt {4}-\sqrt {3})}+... +\frac {\sqrt {100}-\sqrt {99}}{(\sqrt {100}+\sqrt {99})(\sqrt {100}-\sqrt {99})}=\sqrt {2}-1+\sqrt {3}-\sqrt {2}+\sqrt {4}-\sqrt {3}+... +\sqrt {100}-\sqrt {99}=\sqrt {100}-1=10-1=9$.
(3)$\because a=\frac {1}{\sqrt {5}-2}=\frac {\sqrt {5}+2}{(\sqrt {5}-2)(\sqrt {5}+2)}=\sqrt {5}+2$,
$\therefore a-2=\sqrt {5}$.
$\therefore (a-2)^{2}=5$,即$a^{2}-4a+4=5$.
$\therefore a^{2}-4a=1$.
$\therefore 3a^{2}-12a-1=3(a^{2}-4a)-1=3×1-1=2$.
(1)$\frac {1}{\sqrt {2}+1}=\frac {\sqrt {2}-1}{(\sqrt {2}+1)(\sqrt {2}-1)}=\sqrt {2}-1$.
(2)原式$=\frac {\sqrt {2}-1}{(\sqrt {2}+1)(\sqrt {2}-1)}+\frac {\sqrt {3}-\sqrt {2}}{(\sqrt {3}+\sqrt {2})(\sqrt {3}-\sqrt {2})}+\frac {\sqrt {4}-\sqrt {3}}{(\sqrt {4}+\sqrt {3})(\sqrt {4}-\sqrt {3})}+... +\frac {\sqrt {100}-\sqrt {99}}{(\sqrt {100}+\sqrt {99})(\sqrt {100}-\sqrt {99})}=\sqrt {2}-1+\sqrt {3}-\sqrt {2}+\sqrt {4}-\sqrt {3}+... +\sqrt {100}-\sqrt {99}=\sqrt {100}-1=10-1=9$.
(3)$\because a=\frac {1}{\sqrt {5}-2}=\frac {\sqrt {5}+2}{(\sqrt {5}-2)(\sqrt {5}+2)}=\sqrt {5}+2$,
$\therefore a-2=\sqrt {5}$.
$\therefore (a-2)^{2}=5$,即$a^{2}-4a+4=5$.
$\therefore a^{2}-4a=1$.
$\therefore 3a^{2}-12a-1=3(a^{2}-4a)-1=3×1-1=2$.
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