《2025年阳光课堂金牌练习册八年级数学下册人教版福建专版》

1.化简$\sqrt{8}-\sqrt{2}(\sqrt{2}+2)$得( )
A.-2
B.$\sqrt{2}-2$
C.2
D.$4\sqrt{2}-2$

答案： A 解析：$\sqrt{8}-\sqrt{2}(\sqrt{2}+2)=2\sqrt{2}-\sqrt{2}\times\sqrt{2}-2\sqrt{2}=-2$.

2.计算$(2\sqrt{48}-3\sqrt{27})÷\sqrt{6}$的结果是( )
A.$\sqrt{2}$
B.$-\sqrt{2}$
C.$-\frac{\sqrt{2}}{2}$
D.$\frac{\sqrt{2}}{2}$

答案： C 解析：$(2\sqrt{48}-3\sqrt{27}) \div \sqrt{6}=(2\times4\sqrt{3}-3\times3\sqrt{3}) \div \sqrt{6}=-\sqrt{3} \div \sqrt{6}=-\frac{\sqrt{2}}{2}$.

3.$(\sqrt{5}+\sqrt{3})÷\frac{1}{\sqrt{5}-\sqrt{3}}×(\sqrt{5}-\sqrt{3})$等于( )
A.$\sqrt{5}+\sqrt{3}$
B.$\sqrt{5}-\sqrt{3}$
C.$2\sqrt{5}-2\sqrt{3}$
D.$\frac{2}{\sqrt{5}-\sqrt{3}}$

答案： C 解析：$(\sqrt{5}+\sqrt{3}) \div \frac{1}{\sqrt{5}-\sqrt{3}} \times (\sqrt{5}-\sqrt{3})=(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3}) \times (\sqrt{5}-\sqrt{3})=[(\sqrt{5})^{2}-(\sqrt{3})^{2}] \times (\sqrt{5}-\sqrt{3})=2(\sqrt{5}-\sqrt{3})=2\sqrt{5}-2\sqrt{3}$.

4.$(\sqrt{24}-3\sqrt{15}+\sqrt{2\frac{2}{3}})×\sqrt{2}$的值是( )
A.$\frac{16}{3}\sqrt{3}-3\sqrt{30}$
B.$3\sqrt{30}-\frac{2}{3}\sqrt{3}$
C.$2\sqrt{30}-\frac{2}{3}\sqrt{3}$
D.$\frac{20}{3}\sqrt{3}-\sqrt{30}$

答案： A 解析：原式$=\sqrt{24} \times \sqrt{2}-3\sqrt{15} \times \sqrt{2}+\sqrt{2\frac{2}{3}} \times \sqrt{2}=\sqrt{48}-3\sqrt{30}+\sqrt{\frac{16}{3}}=4\sqrt{3}-3\sqrt{30}+\frac{4\sqrt{3}}{3}=(4\sqrt{3}+\frac{4\sqrt{3}}{3})-3\sqrt{30}=\frac{16\sqrt{3}}{3}-3\sqrt{30}$.

5.若$a=\frac{\sqrt{5}+\sqrt{3}}{2}$，$b=\frac{\sqrt{5}-\sqrt{3}}{2}$，则$a^{2}-ab+b^{2}$的值为( )
A.$\frac{7}{2}$
B.$\frac{9}{2}$
C.$\frac{11}{2}$
D.$\frac{\sqrt{15}-2}{2}$

答案： A 解析：$a^{2}-ab + b^{2}=(a + b)^{2}-3ab=(\frac{\sqrt{5}+\sqrt{3}}{2}+\frac{\sqrt{5}-\sqrt{3}}{2})^{2}-3\times\frac{\sqrt{5}+\sqrt{3}}{2} \times \frac{\sqrt{5}-\sqrt{3}}{2}=5-\frac{3}{2}=\frac{7}{2}$.

6.若$x=\sqrt{2}+1$，则$x^{2}-2x+1$等于( )
A.$\sqrt{2}$
B.$\sqrt{2}-1$
C.$2+\sqrt{2}$
D.2

答案： D 解析：$x^{2}-2x + 1=(x - 1)^{2}=(\sqrt{2}+1 - 1)^{2}=(\sqrt{2})^{2}=2$.

7.若$\sqrt{x}+\sqrt{y}=\sqrt{5}+\sqrt{3}$，$\sqrt{xy}=\sqrt{15}-\sqrt{3}$，则$x + y =$________.

答案： $8 + 2\sqrt{3}$ 解析：由$(\sqrt{x}+\sqrt{y})^{2}=(\sqrt{5}+\sqrt{3})^{2}$，得$x + y+2\sqrt{xy}=8 + 2\sqrt{15}$. 又因为$\sqrt{xy}=\sqrt{15}-\sqrt{3}$，所以$x + y+2(\sqrt{15}-\sqrt{3})=8 + 2\sqrt{15}$，所以$x + y=8 + 2\sqrt{3}$.

8.已知$a = 2+\sqrt{3}$，$b = 2-\sqrt{3}$，那么$\frac{1}{a}+\frac{1}{b}=$________.

答案： 4 解析：$\frac{1}{a}+\frac{1}{b}=\frac{1}{2+\sqrt{3}}+\frac{1}{2-\sqrt{3}}=2-\sqrt{3}+2+\sqrt{3}=4$.

9.计算：$(\sqrt{2}+\sqrt{3}-\sqrt{10})(\sqrt{2}-\sqrt{3}+\sqrt{10})=$________.

答案： $2\sqrt{30}-11$ 解析：原式$=(\sqrt{2}+\sqrt{3}-\sqrt{10})[\sqrt{2}-(\sqrt{3}-\sqrt{10})]=(\sqrt{2})^{2}-(\sqrt{3}-\sqrt{10})^{2}=2-(3 + 10-2\sqrt{30})=2\sqrt{30}-11$.

10.$(2\sqrt{2}-3)^{3024}(2\sqrt{2}+3)^{3024}=$________.

答案： 1 解析：原式$=[(2\sqrt{2})^{2}-3^{2}]^{3024}=(8 - 9)^{3024}=1$.

11.计算：(1)$\sqrt{2}×(\sqrt{2}+\frac{1}{\sqrt{2}})-\frac{\sqrt{18}-\sqrt{8}}{2}$；
(2)$(2\sqrt{3}-\sqrt{5})^{2}$.

答案： 解：
(1)原式$=(\sqrt{2})^{2}+\sqrt{2} \times \frac{1}{\sqrt{2}}-\frac{3\sqrt{2}-2\sqrt{2}}{2}=2 + 1-\frac{\sqrt{2}}{2}=3-\frac{\sqrt{2}}{2}$.
(2)原式$=(2\sqrt{3})^{2}-2\times2\sqrt{3} \times \sqrt{5}+(\sqrt{5})^{2}=12-4\sqrt{15}+5=17-4\sqrt{15}$.

12.已知$a = 3+2\sqrt{2}$，$b = 3-2\sqrt{2}$，求$a^{2}b - ab^{2}$的值.

答案： 解：因为$a = 3 + 2\sqrt{2}$，$b = 3 - 2\sqrt{2}$，所以$ab=(3 + 2\sqrt{2})(3 - 2\sqrt{2})=1$，$a - b=3 + 2\sqrt{2}-3 + 2\sqrt{2}=4\sqrt{2}$. 所以$a^{2}b - ab^{2}=ab(a - b)=4\sqrt{2}$.