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11. 若$x^{2}-x-2=0$，则代数式$\dfrac{x^{2}-x+2\sqrt{3}}{(x^{2}-x)^{2}-1+\sqrt{3}}$的值为(
A.$\dfrac{2\sqrt{3}}{3}$
B.$\dfrac{\sqrt{3}}{3}$
C.$\sqrt{3}$
D.$\sqrt{3}$或$\dfrac{\sqrt{3}}{3}$
A
)A.$\dfrac{2\sqrt{3}}{3}$
B.$\dfrac{\sqrt{3}}{3}$
C.$\sqrt{3}$
D.$\sqrt{3}$或$\dfrac{\sqrt{3}}{3}$
答案
11. A
12. 计算$(\sqrt{3}+\sqrt{2})^{2}(\sqrt{3}-\sqrt{2})$的结果是
$\sqrt{3}+\sqrt{2}$
。答案
12. $\sqrt{3}+\sqrt{2}$
13. 已知$x=\dfrac{1}{2}(\sqrt{7}+\sqrt{5})$，$y=\dfrac{1}{2}(\sqrt{7}-\sqrt{5})$。
(1)求$x + y$的值和$xy$的值；
(2)求$\dfrac{y}{x}+\dfrac{x}{y}$的值。
(1)求$x + y$的值和$xy$的值；
(2)求$\dfrac{y}{x}+\dfrac{x}{y}$的值。
答案
13. 解：（1）$\because x=\frac{1}{2}(\sqrt{7}+\sqrt{5})$，$y=\frac{1}{2}(\sqrt{7}-\sqrt{5})$，
$\therefore x+y=\frac{1}{2}(\sqrt{7}+\sqrt{5})+\frac{1}{2}(\sqrt{7}-\sqrt{5})=\frac{1}{2}(\sqrt{7}+\sqrt{5}+\sqrt{7}-\sqrt{5})=\frac{1}{2}× 2\sqrt{7}=\sqrt{7}$，$xy=\frac{1}{2}(\sqrt{7}+\sqrt{5})× \frac{1}{2}(\sqrt{7}-\sqrt{5})=\frac{1}{4}× [(\sqrt{7})^{2}-(\sqrt{5})^{2}]=\frac{1}{4}× 2=\frac{1}{2}$.
（2）$\because x=\frac{1}{2}(\sqrt{7}+\sqrt{5})$，$y=\frac{1}{2}(\sqrt{7}-\sqrt{5})$，
$\therefore \frac{y}{x}+\frac{x}{y}=\frac{\frac{1}{2}(\sqrt{7}-\sqrt{5})}{\frac{1}{2}(\sqrt{7}+\sqrt{5})}+\frac{\frac{1}{2}(\sqrt{7}+\sqrt{5})}{\frac{1}{2}(\sqrt{7}-\sqrt{5})}=\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}+\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}=\frac{(\sqrt{7}-\sqrt{5})^{2}}{2}+\frac{(\sqrt{7}+\sqrt{5})^{2}}{2}=\frac{7-2\sqrt{35}+5+7+2\sqrt{35}+5}{2}=\frac{24}{2}=12$.
$\therefore x+y=\frac{1}{2}(\sqrt{7}+\sqrt{5})+\frac{1}{2}(\sqrt{7}-\sqrt{5})=\frac{1}{2}(\sqrt{7}+\sqrt{5}+\sqrt{7}-\sqrt{5})=\frac{1}{2}× 2\sqrt{7}=\sqrt{7}$，$xy=\frac{1}{2}(\sqrt{7}+\sqrt{5})× \frac{1}{2}(\sqrt{7}-\sqrt{5})=\frac{1}{4}× [(\sqrt{7})^{2}-(\sqrt{5})^{2}]=\frac{1}{4}× 2=\frac{1}{2}$.
（2）$\because x=\frac{1}{2}(\sqrt{7}+\sqrt{5})$，$y=\frac{1}{2}(\sqrt{7}-\sqrt{5})$，
$\therefore \frac{y}{x}+\frac{x}{y}=\frac{\frac{1}{2}(\sqrt{7}-\sqrt{5})}{\frac{1}{2}(\sqrt{7}+\sqrt{5})}+\frac{\frac{1}{2}(\sqrt{7}+\sqrt{5})}{\frac{1}{2}(\sqrt{7}-\sqrt{5})}=\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}+\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}=\frac{(\sqrt{7}-\sqrt{5})^{2}}{2}+\frac{(\sqrt{7}+\sqrt{5})^{2}}{2}=\frac{7-2\sqrt{35}+5+7+2\sqrt{35}+5}{2}=\frac{24}{2}=12$.
14. 观察下列运算过程：
$\dfrac{1}{1+\sqrt{2}}=\dfrac{1}{\sqrt{2}+1}=\dfrac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\dfrac{\sqrt{2}-1}{(\sqrt{2})^{2}-1^{2}}=\sqrt{2}-1$；
$\dfrac{1}{\sqrt{2}+\sqrt{3}}=\dfrac{1}{\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}=\sqrt{3}-\sqrt{2}······$
请运用上面的运算方法计算：$\dfrac{1}{1+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+···+\dfrac{1}{\sqrt{2025}+\sqrt{2027}}+\dfrac{1}{\sqrt{2027}+\sqrt{2029}}=$
$\dfrac{1}{1+\sqrt{2}}=\dfrac{1}{\sqrt{2}+1}=\dfrac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}=\dfrac{\sqrt{2}-1}{(\sqrt{2})^{2}-1^{2}}=\sqrt{2}-1$；
$\dfrac{1}{\sqrt{2}+\sqrt{3}}=\dfrac{1}{\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}=\sqrt{3}-\sqrt{2}······$
请运用上面的运算方法计算：$\dfrac{1}{1+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+···+\dfrac{1}{\sqrt{2025}+\sqrt{2027}}+\dfrac{1}{\sqrt{2027}+\sqrt{2029}}=$
$\frac{\sqrt{2029}-1}{2}$
。答案
14. $\frac{\sqrt{2029}-1}{2}$
