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4.已知$x= 2-\sqrt {3}$,则代数式$(7+4\sqrt {3})x^{2}+(2+\sqrt {3})x+\sqrt {3}$的值是 ()
A.0
B.$\sqrt {3}$
C.$2+\sqrt {3}$
D.$2-\sqrt {3}$
A.0
B.$\sqrt {3}$
C.$2+\sqrt {3}$
D.$2-\sqrt {3}$
答案
C
5.已知$a= \sqrt {5}+2,b= 2-\sqrt {5}$,则$a^{2025}b^{2024}$的值为______.
答案
$\sqrt {5}+2$
6.计算:
(1)$(\sqrt {3}-\sqrt {2})^{2}-(\sqrt {3}+\sqrt {2})^{2}$;
(2)$(\sqrt {5}+\sqrt {2})(\sqrt {5}-\sqrt {2})-(\sqrt {3}-\sqrt {2})^{2}$.
(1)$(\sqrt {3}-\sqrt {2})^{2}-(\sqrt {3}+\sqrt {2})^{2}$;
(2)$(\sqrt {5}+\sqrt {2})(\sqrt {5}-\sqrt {2})-(\sqrt {3}-\sqrt {2})^{2}$.
答案
(1)$-4\sqrt {6}$ (2)$-2+2\sqrt {6}$
7.已知$a^{2}-\sqrt {13}a+1= 0$,求$a-\frac {1}{a}$的值.
答案
$\because a≠0$,∴在已知等式两边都除以a,得$a-\sqrt {13}+\frac {1}{a}=0,\therefore a+\frac {1}{a}=\sqrt {13},\therefore (a-\frac {1}{a})^{2}=(a+\frac {1}{a})^{2}-4=(\sqrt {13})^{2}-4=13-4=9,\therefore a-\frac {1}{a}=\pm 3$
8.已知$a+b= -8,ab= 8$,则$b\sqrt {\frac {a}{b}}+a\sqrt {\frac {b}{a}}$的值为______.
答案
$-4\sqrt {2}$
9.已知$x= \frac {1}{2}(\sqrt {5}+\sqrt {3}),y= \frac {1}{2}(\sqrt {5}-\sqrt {3})$,求下列各式的值:
(1)$x^{2}-xy+y^{2}$;
(2)$\frac {x}{y}+\frac {y}{x}$.
(1)$x^{2}-xy+y^{2}$;
(2)$\frac {x}{y}+\frac {y}{x}$.
答案
$x+y=\frac {1}{2}(\sqrt {5}+\sqrt {3})+\frac {1}{2}(\sqrt {5}-\sqrt {3})=\sqrt {5},xy=\frac {1}{2}(\sqrt {5}+\sqrt {3})×\frac {1}{2}(\sqrt {5}-\sqrt {3})=\frac {1}{2}$ (1)原式$=(x+y)^{2}-3xy=(\sqrt {5})^{2}-\frac {3}{2}=5-\frac {3}{2}=\frac {7}{2}$ (2)原式$=\frac {x^{2}+y^{2}}{xy}=\frac {(x+y)^{2}-2xy}{xy}=\frac {5-1}{\frac {1}{2}}=8$
