《2025年暑假总动员八年级数学沪科版合肥工业大学出版社》

19. 若$a$,$b$均为有理数,且$\sqrt{8}+\sqrt{18}+\sqrt{\dfrac{1}{8}}= (a + b)\sqrt{2}$,则$a + b = $$\frac{21}{4}$.

答案： $\frac{21}{4}$

20. 计算:
(1)$\sqrt{18}-\sqrt{\dfrac{1}{2}}÷\sqrt{\dfrac{4}{3}}×\dfrac{6}{\sqrt{3}}$;
(2)$(\sqrt{10}+\sqrt{7})(\sqrt{10}-\sqrt{7})-(\sqrt{2}+1)^{2}$;
(3)$\left(2\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{1}{2}}\right)\left(\dfrac{1}{2}\sqrt{8}+\sqrt{\dfrac{2}{3}}\right)$.

答案： 解：
(1)原式 $= 3\sqrt{2} - \frac{1}{\sqrt{2}} \div \frac{\sqrt{4}}{\sqrt{3}} \times \frac{6}{\sqrt{3}} = 3\sqrt{2} -\frac{\sqrt{2}}{2} \div \frac{2\sqrt{3}}{3} \times \frac{6}{\sqrt{3}} = 3\sqrt{2} - \frac{\sqrt{2}}{2} \times \frac{3}{2\sqrt{3}} \times \frac{6}{\sqrt{3}} = 3\sqrt{2} -\frac{3}{2}\sqrt{2} = \frac{3}{2}\sqrt{2}$.
(2)原式 $= 10 - 7 - (2 + 2\sqrt{2} + 1) = -2\sqrt{2}$.
(3)原式 $= 2\sqrt{\frac{3}{2}} \times \frac{1}{2}\sqrt{8} + 2\sqrt{\frac{3}{2}} \times \sqrt{\frac{2}{3}} - \sqrt{\frac{1}{2}} \times\frac{1}{2}\sqrt{8} - \sqrt{\frac{1}{2}} \times \sqrt{\frac{2}{3}} = 2 \times \frac{1}{2}\sqrt{\frac{3}{2} \times 8} + 2\sqrt{\frac{3}{2} \times \frac{2}{3}} -\frac{1}{2}\sqrt{\frac{1}{2} \times 8} - \sqrt{\frac{1}{2} \times \frac{2}{3}} = \sqrt{12} + 2 - \frac{1}{2}\sqrt{4} - \sqrt{\frac{1}{3}} =2\sqrt{3} + 2 - 1 - \frac{\sqrt{3}}{3} = (2 - \frac{1}{3})\sqrt{3} + 1 = \frac{5\sqrt{3}}{3} + 1$.

21. 已知$a = \dfrac{\sqrt{3}-1}{\sqrt{3}+1}$,$b = \dfrac{\sqrt{3}+1}{\sqrt{3}-1}$,求$a^{3}+b^{3}-4$的值.
解：$\because a = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}, b = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}, \therefore a + b = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} + \frac{\sqrt{3} + 1}{\sqrt{3} - 1} =\frac{(\sqrt{3} - 1)^2 + (\sqrt{3} + 1)^2}{2} = \frac{3 - 2\sqrt{3} + 1 + 3 + 2\sqrt{3} + 1}{2} = \frac{8}{2} = 4$,$ab = 1, \therefore (a + b)^2 = 16, \therefore a^2 + 2ab + b^2 = 16$,$\therefore a^2 + b^2 = 16 - 2ab = 16 - 2 = 14, \therefore a^3 + b^3 -4 = (a + b)(a^2 - ab + b^2) - 4 = 4 × (14 - 1) - 4 =52 - 4 = 48$.

答案： 解：$\because a = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}, b = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}, \therefore a + b = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} + \frac{\sqrt{3} + 1}{\sqrt{3} - 1} =\frac{(\sqrt{3} - 1)^2 + (\sqrt{3} + 1)^2}{2} = \frac{3 - 2\sqrt{3} + 1 + 3 + 2\sqrt{3} + 1}{2} = \frac{8}{2} = 4$,$ab = 1, \therefore (a + b)^2 = 16, \therefore a^2 + 2ab + b^2 = 16$,$\therefore a^2 + b^2 = 16 - 2ab = 16 - 2 = 14, \therefore a^3 + b^3 -4 = (a + b)(a^2 - ab + b^2) - 4 = 4 \times (14 - 1) - 4 =52 - 4 = 48$.

22. 若$a = \sqrt{2}-3$,求$2(a - \sqrt{2})+(a+\sqrt{2})-a(a - 3)+4$的值.

答案： 解：原式 $= 2a - 2\sqrt{2} + a + \sqrt{2} - a^2 + 3a + 4 =-a^2 + 6a + 4 - \sqrt{2}$. 当 $a = \sqrt{2} - 3$ 时，原式 $=-(\sqrt{2} - 3)^2 + 6(\sqrt{2} - 3) + 4 - \sqrt{2} = 11\sqrt{2} - 25$.

23. 观察下列各式及其验证过程.
$2\sqrt{\dfrac{2}{3}}= \sqrt{2+\dfrac{2}{3}}$
验证方法1:$2\sqrt{\dfrac{2}{3}}= \sqrt{\dfrac{2^{3}}{3}}= \sqrt{\dfrac{(2^{3}-2)+2}{2^{2}-1}}= \sqrt{\dfrac{2(2^{2}-1)+2}{2^{2}-1}}= \sqrt{2+\dfrac{2}{3}}$.
验证方法2:$2\sqrt{\dfrac{2}{3}}= \sqrt{\dfrac{2^{3}}{3}}= \sqrt{\dfrac{8}{3}}= \sqrt{2+\dfrac{2}{3}}$.
$3\sqrt{\dfrac{3}{8}}= \sqrt{3+\dfrac{3}{8}}$
验证方法1:$3\sqrt{\dfrac{3}{8}}= \sqrt{\dfrac{3^{3}}{8}}= \sqrt{\dfrac{(3^{3}-3)+3}{3^{2}-1}}= \sqrt{\dfrac{3(3^{2}-1)+3}{3^{2}-1}}= \sqrt{3+\dfrac{3}{8}}$.
验证方法2:$3\sqrt{\dfrac{3}{8}}= \sqrt{\dfrac{3^{3}}{8}}= \sqrt{\dfrac{27}{8}}= \sqrt{3+\dfrac{3}{8}}$.
(1)针对上述两个等式及其验证过程的基本思路,猜想$4\sqrt{\dfrac{4}{15}}$的变形结果,并进行验证;猜想：$4\sqrt{\frac{4}{15}} = \sqrt{4 + \frac{4}{15}}$，验证方法 1：$4\sqrt{\frac{4}{15}} = \sqrt{\frac{4^3}{15}} = \sqrt{\frac{(4^3 - 4) + 4}{4^2 - 1}} = \sqrt{\frac{4(4^2 - 1) + 4}{4^2 - 1}} =\sqrt{4 + \frac{4}{15}}$. 验证方法 2：$4\sqrt{\frac{4}{15}} = \sqrt{\frac{4^3}{15}} = \sqrt{\frac{64}{15}} =\sqrt{4 + \frac{4}{15}}$.
(2)针对上述各等式反映的规律,写出用$n$($n\geqslant 2且n$为自然数)表示的等式;$n\sqrt{\frac{n}{n^2 - 1}} = \sqrt{n + \frac{n}{n^2 - 1}}(n \geq 2$ 且 $n$ 为自然数).
(3)在上述等式的验证过程中,用到等式$\sqrt{n^{2}} = n(n\geqslant 0)$,能否得到等式$\sqrt[3]{n^{3}} = n(n\geqslant 0)$,$\sqrt[4]{n^{4}} = n(n\geqslant 0)$?其一般规律可以表示成什么样的等式?请写出一般式.能得到，一般式为$\sqrt[k]{n^k} = n(n \geq 0, k$ 为正整数).

答案： 解：
(1)猜想：$4\sqrt{\frac{4}{15}} = \sqrt{4 + \frac{4}{15}}$，验证方法 1：$4\sqrt{\frac{4}{15}} = \sqrt{\frac{4^3}{15}} = \sqrt{\frac{(4^3 - 4) + 4}{4^2 - 1}} = \sqrt{\frac{4(4^2 - 1) + 4}{4^2 - 1}} =\sqrt{4 + \frac{4}{15}}$. 验证方法 2：$4\sqrt{\frac{4}{15}} = \sqrt{\frac{4^3}{15}} = \sqrt{\frac{64}{15}} =\sqrt{4 + \frac{4}{15}}$.
(2)$n\sqrt{\frac{n}{n^2 - 1}} = \sqrt{n + \frac{n}{n^2 - 1}}(n \geq 2$ 且 $n$ 为自然数).
(3)能得到，一般式为 $\sqrt[k]{n^k} = n(n \geq 0, k$ 为正整数).