《2025年星推荐涂教材八年级数学下册人教版》

4.（天津中考）计算$(\sqrt{7} + \sqrt{6})(\sqrt{7} - \sqrt{6})$的结果为______.

答案： [名师讲习]原式$ = (\sqrt{7})^2 - (\sqrt{6})^2 = 7 - 6 = 1$. [正确答案]1

5.（2024·武威中考）计算：$\sqrt{18} - \sqrt{12} \times \sqrt{\frac{3}{2}}$.

答案： [解]$\sqrt{18} - \sqrt{12} \times \sqrt{\frac{3}{2}} = \sqrt{18} - \sqrt{12 \times \frac{3}{2}} = \sqrt{18} - \sqrt{18} = 0$.

[针对性训练]
4.（聊城中考）计算：$(\sqrt{48} - 3\sqrt{\frac{1}{3}}) \div \sqrt{3} =$______.

答案： [解]$(\sqrt{48} - 3\sqrt{\frac{1}{3}}) \div \sqrt{3} = (\sqrt{16 \times 3} - 3 \times \frac{\sqrt{3}}{3}) \div \sqrt{3} = (4\sqrt{3} - \sqrt{3}) \div \sqrt{3} = 3\sqrt{3} \div \sqrt{3} = 3$. [正确答案]3

6.（济宁中考）已知$a = 2 + \sqrt{5}$，$b = 2 - \sqrt{5}$，求代数式$a^2b + ab^2$的值.

答案： [解]$\because a = 2 + \sqrt{5}$，$b = 2 - \sqrt{5}$，$\therefore a^2b + ab^2 = ab(a + b) = (2 + \sqrt{5}) \times (2 - \sqrt{5}) \times (2 + \sqrt{5} + 2 - \sqrt{5}) = (4 - 5) \times 4 = -1 \times 4 = -4$.

[针对性训练]
5.（青海中考）先化简，再求值：$(a - \frac{1}{a}) \div \frac{a^2 - 2a + 1}{a}$，其中$a = \sqrt{2} + 1$.

答案：  [解]$(a - \frac{1}{a}) \div \frac{a^2 - 2a + 1}{a} = \frac{a^2 - 1}{a} \div \frac{(a - 1)^2}{a} = \frac{(a + 1)(a - 1)}{a} \cdot \frac{a}{(a - 1)^2} = \frac{a + 1}{a - 1}$.当$a = \sqrt{2} + 1$时，原式$ = \frac{\sqrt{2} + 1 + 1}{\sqrt{2} + 1 - 1} = \frac{\sqrt{2} + 2}{\sqrt{2}} = 1 + \sqrt{2}$.