《2025年通城学典活页检测八年级数学下册浙教版》

12. (16分)计算：
(1) $\sqrt{48}\div\sqrt{3}-\sqrt{\frac{1}{2}}\times\sqrt{12}+\sqrt{24}$； (2) $(\sqrt{7}+\sqrt{5})\times(\sqrt{28}-\sqrt{20})-(\sqrt{3}+3\sqrt{2})^{2}$.

答案：
(1)原式$=4-\sqrt{6}+2\sqrt{6}=4+\sqrt{6}$
(2)原式$=(\sqrt{7}+\sqrt{5})\times(2\sqrt{7}-2\sqrt{5})-(3 + 6\sqrt{6}+18)=2(\sqrt{7}+\sqrt{5})\times(\sqrt{7}-\sqrt{5})-21-6\sqrt{6}=2\times(7 - 5)-21-6\sqrt{6}=-17-6\sqrt{6}$

13. (10分)已知$x=\sqrt{5}-2$，求$(9 + 4\sqrt{5})x^{2}-(\sqrt{5}+2)x + 4$的值.

答案： $\because x=\sqrt{5}-2$，$\therefore(9 + 4\sqrt{5})x^{2}-(\sqrt{5}+2)x + 4=(9 + 4\sqrt{5})\times(\sqrt{5}-2)^{2}-(\sqrt{5}+2)\times(\sqrt{5}-2)+4=(9 + 4\sqrt{5})\times(9 - 4\sqrt{5})-1+4=81 - 80-1+4=4$

14. (13分)不用计算器，比较$\frac{\sqrt{30}}{\sqrt{6}-\sqrt{3}}$与$2\sqrt{5}+\sqrt{3}$的大小，并说明理由.

答案： $\frac{\sqrt{30}}{\sqrt{6}-\sqrt{3}}＞2\sqrt{5}+\sqrt{3}$ 理由：$\because\frac{\sqrt{30}}{\sqrt{6}-\sqrt{3}}=\frac{\sqrt{30}(\sqrt{6}+\sqrt{3})}{(\sqrt{6}-\sqrt{3})(\sqrt{6}+\sqrt{3})}=\frac{6\sqrt{5}+3\sqrt{10}}{3}=2\sqrt{5}+\sqrt{10}$，$\sqrt{10}＞\sqrt{3}$，$\therefore 2\sqrt{5}+\sqrt{10}＞2\sqrt{5}+\sqrt{3}$，$\therefore\frac{\sqrt{30}}{\sqrt{6}-\sqrt{3}}＞2\sqrt{5}+\sqrt{3}$