猿辅导2017年春季初联训练营作业题解答-3: "二次根式"

 

1. $\sqrt{16 - 2\sqrt{20} - 2\sqrt{28} + 2\sqrt{35}}$

解答:

仿例题9可得: $$\sqrt{16 - 2\sqrt{20} - 2\sqrt{28} + 2\sqrt{35}} = \sqrt{(\sqrt7 + \sqrt5 - 2)^2} = \sqrt7 + \sqrt5 - 2.$$

 

 

2. $\sqrt{12 + \sqrt{24} + \sqrt{39} + \sqrt{104}} - \sqrt{12 - \sqrt{24} + \sqrt{39} - \sqrt{104}}$

解答:

仿例题4可得: $$\begin{cases}x^2 + y^2 = 24 + 2\sqrt{39}\\ xy = \sqrt{55+8\sqrt{39}} = \sqrt{39} + 4\end{cases}$$ $$\Rightarrow (x - y)^2 = 16 \Rightarrow x - y = 4.$$

 

 

3. $\sqrt{2a - \sqrt{3a^2 - 2ab - b^2}}$ ($a > b > 0$)

解答:

仿例题5可得: $$\sqrt{2a - \sqrt{3a^2 - 2ab - b^2}} = \sqrt{2a - \sqrt{(3a + b)(a - b)}} = \sqrt{\frac{1}{2}\left(4a - 2\sqrt{(3a+b)(a-b)}\right)}$$ $$= \frac{\sqrt2}{2}\left(\sqrt{3a+b} - \sqrt{a - b}\right).$$

 

 

4. $\sqrt{x + 2 + 3\sqrt{2x-5}} - \sqrt{x - 2 + \sqrt{2x - 5}}$ $\left(x \ge \dfrac{5}{2}\right)$

解答: $$\sqrt{x + 2 + 3\sqrt{2x-5}} - \sqrt{x - 2 + \sqrt{2x - 5}}$$ $$= \sqrt{\frac{1}{2}\left(2x + 4 + 6\sqrt{2x - 5}\right)} - \sqrt{\frac{1}{2}\left(2x - 4 + 2\sqrt{2x - 5}\right)}$$ $$= \sqrt{\frac{1}{2}\left(\sqrt{2x-5} + 3\right)^2} - \sqrt{\frac{1}{2}\left(\sqrt{2x - 5} + 1\right)^2}$$ $$= \frac{\sqrt2}{2}\left(\sqrt{2x-5} + 3\right) - \frac{\sqrt2}{2}\left(\sqrt{2x - 5} + 1\right) = \sqrt2.$$

 

 

5. $\dfrac{\left(11+6\sqrt2\right)\sqrt{11-6\sqrt2} - \left(11-6\sqrt2\right)\sqrt{11+6\sqrt2}}{\left(\sqrt{\sqrt5 + 2}+ \sqrt{\sqrt5 - 2}\right) \div \sqrt{\sqrt5 + 1}}$

解答: $$11 \pm 6\sqrt2 = \left(3 \pm \sqrt2\right)^2,$$ $$\left(\sqrt{\sqrt5 + 2} + \sqrt{\sqrt5 - 2}\right)^2 = 2\sqrt5 + 2 = 2\left(\sqrt5 + 1\right).$$ $$\Rightarrow \frac{\left(3 + \sqrt2\right)^2\left(3 - \sqrt2\right) - \left(3 - \sqrt2\right)^2\left(3 + \sqrt2\right)}{\sqrt2}$$ $$= \frac{(3 + \sqrt2)(3 - \sqrt2)(3+\sqrt2 - 3 + \sqrt2)}{\sqrt2} = 14.$$

 

 

6. $\dfrac{2+2\sqrt7 +\sqrt{10}}{\left(\sqrt7 + \sqrt{10}\right)\left(\sqrt7 + 2\right)} + \dfrac{4+2\sqrt{13} + \sqrt{10}}{\left(\sqrt{10} + \sqrt{13}\right)\left(\sqrt{13} + 4\right)}$

解答: $$\frac{1}{\sqrt7 + \sqrt{10}} + \frac{1}{\sqrt{7} + 2} + \frac{1}{\sqrt{10} + \sqrt{13}} + \frac{1}{\sqrt{13} + 4}$$ $$= \frac{1}{3}\left(\sqrt{10} - \sqrt7 + \sqrt7 - 2 + \sqrt{13} - \sqrt{10} + 4 - \sqrt{13}\right) = \frac{2}{3}.$$

 

7. $\dfrac{9 + 2\sqrt6 + 2\sqrt{14} + \sqrt{21}}{2\sqrt2 + \sqrt3 + \sqrt7}$

解答: $$\left(2\sqrt2 + \sqrt3 + \sqrt7\right)^2 = 18 + 4\sqrt6 + 4\sqrt{14} + 2\sqrt{21}$$ $$= 2\left(9 + 2\sqrt6+ 2\sqrt{14} + 2\sqrt{21}\right).$$ $$\Rightarrow \frac{\dfrac{1}{2}\left(2\sqrt2 + \sqrt3 + \sqrt7\right)^2}{2\sqrt2 + \sqrt3 + \sqrt7}= \frac{1}{2}\left(2\sqrt2 + \sqrt3 + \sqrt7\right).$$

 

 

主讲教师:

赵胤, 理学硕士(数学) & 教育硕士(数学), 中国数学奥林匹克一级教练员, 高级中学数学教师资格.