登录
11. 已知$a + b = - 3$，$ab = 2$，求代数式$\sqrt{\frac{b}{a}} + \sqrt{\frac{a}{b}}$的值.
答案
11. $\dfrac{3\sqrt{2}}{2}$
解析
$\because a + b = - 3$，$ab = 2$，
$\therefore a ＜ 0$，$b ＜ 0$，
$\sqrt{\frac{b}{a}} + \sqrt{\frac{a}{b}} = \sqrt{\frac{ab}{a^2}} + \sqrt{\frac{ab}{b^2}} = \frac{\sqrt{ab}}{|a|} + \frac{\sqrt{ab}}{|b|} = -\frac{\sqrt{ab}}{a} - \frac{\sqrt{ab}}{b} = -\sqrt{ab}(\frac{1}{a} + \frac{1}{b}) = -\sqrt{ab}·\frac{a + b}{ab}$，
将$a + b = - 3$，$ab = 2$代入，得$-\sqrt{2}·\frac{- 3}{2} = \frac{3\sqrt{2}}{2}$.
$\therefore a ＜ 0$，$b ＜ 0$，
$\sqrt{\frac{b}{a}} + \sqrt{\frac{a}{b}} = \sqrt{\frac{ab}{a^2}} + \sqrt{\frac{ab}{b^2}} = \frac{\sqrt{ab}}{|a|} + \frac{\sqrt{ab}}{|b|} = -\frac{\sqrt{ab}}{a} - \frac{\sqrt{ab}}{b} = -\sqrt{ab}(\frac{1}{a} + \frac{1}{b}) = -\sqrt{ab}·\frac{a + b}{ab}$，
将$a + b = - 3$，$ab = 2$代入，得$-\sqrt{2}·\frac{- 3}{2} = \frac{3\sqrt{2}}{2}$.
12. 当$a = - 3$，$b = - 2$时，求$\frac{b}{a - 2b}\sqrt{\frac{a^3 - 4a^2b + 4ab^2}{b}}$的值.
答案
12. $-\sqrt{6}$
解析
当$a = - 3$，$b = - 2$时，
$\begin{aligned}&\frac{b}{a - 2b}\sqrt{\frac{a^3 - 4a^2b + 4ab^2}{b}}\\=&\frac{b}{a - 2b}\sqrt{\frac{a(a^2 - 4ab + 4b^2)}{b}}\\=&\frac{b}{a - 2b}\sqrt{\frac{a(a - 2b)^2}{b}}\\=&\frac{b}{a - 2b}·\frac{|a - 2b|\sqrt{ab}}{|b|}\\\end{aligned}$
因为$a=-3$，$b=-2$，所以$a - 2b=-3 - 2×(-2)=-3 + 4=1＞0$，$b=-2＜0$，
$\begin{aligned}&\frac{b}{a - 2b}·\frac{|a - 2b|\sqrt{ab}}{|b|}\\=&\frac{-2}{1}·\frac{1×\sqrt{(-3)×(-2)}}{2}\\=&-2·\frac{\sqrt{6}}{2}\\=&-\sqrt{6}\end{aligned}$
$-\sqrt{6}$
$\begin{aligned}&\frac{b}{a - 2b}\sqrt{\frac{a^3 - 4a^2b + 4ab^2}{b}}\\=&\frac{b}{a - 2b}\sqrt{\frac{a(a^2 - 4ab + 4b^2)}{b}}\\=&\frac{b}{a - 2b}\sqrt{\frac{a(a - 2b)^2}{b}}\\=&\frac{b}{a - 2b}·\frac{|a - 2b|\sqrt{ab}}{|b|}\\\end{aligned}$
因为$a=-3$，$b=-2$，所以$a - 2b=-3 - 2×(-2)=-3 + 4=1＞0$，$b=-2＜0$，
$\begin{aligned}&\frac{b}{a - 2b}·\frac{|a - 2b|\sqrt{ab}}{|b|}\\=&\frac{-2}{1}·\frac{1×\sqrt{(-3)×(-2)}}{2}\\=&-2·\frac{\sqrt{6}}{2}\\=&-\sqrt{6}\end{aligned}$
$-\sqrt{6}$
13. 试按下面两种方法完成计算，并比较哪一种解法简便.
(1)$\frac{1}{\sqrt{2}} \approx \frac{1}{1.414} \approx$
$\frac{1}{\sqrt{2}} =\frac{1 · \sqrt{2}}{\sqrt{2} · \sqrt{2}} =\frac{\sqrt{2}}{2} \approx \frac{1.414}{2} \approx$
(2)在下列各题的横线上填上最简单的二次根式，使它们的积不含根号：
①$\sqrt{3} ·$
②$2\sqrt{6} ·$
③$\sqrt{32} ·$
④$\sqrt{2a^2} ·$
⑤$\sqrt{8x^3} ·$
⑥$\sqrt{x - 1} ·$
(3)根据以上问题解答过程中所得到的启发求下列各式的值(精确到$0.01$)：
①$\frac{6}{\sqrt{3}}$；
②$\frac{2\sqrt{10}}{\sqrt{5}}$；
③$\frac{\sqrt{63}}{2\sqrt{14}}$；
④$\sqrt{1.5}$.
(1)$\frac{1}{\sqrt{2}} \approx \frac{1}{1.414} \approx$
0.71
(精确到$0.01$)，$\frac{1}{\sqrt{2}} =\frac{1 · \sqrt{2}}{\sqrt{2} · \sqrt{2}} =\frac{\sqrt{2}}{2} \approx \frac{1.414}{2} \approx$
0.71
(精确到$0.01$)；(2)在下列各题的横线上填上最简单的二次根式，使它们的积不含根号：
①$\sqrt{3} ·$
$\sqrt{3}$
；②$2\sqrt{6} ·$
$\sqrt{6}$
；③$\sqrt{32} ·$
$\sqrt{2}$
；④$\sqrt{2a^2} ·$
$\sqrt{2}$
；⑤$\sqrt{8x^3} ·$
$\sqrt{2x}$
；⑥$\sqrt{x - 1} ·$
$\sqrt{x - 1}$
；(3)根据以上问题解答过程中所得到的启发求下列各式的值(精确到$0.01$)：
①$\frac{6}{\sqrt{3}}$；
②$\frac{2\sqrt{10}}{\sqrt{5}}$；
③$\frac{\sqrt{63}}{2\sqrt{14}}$；
④$\sqrt{1.5}$.
答案
13. (1) 0.71 0.71 (2) ① $\sqrt{3}$ ② $\sqrt{6}$ ③ $\sqrt{2}$ ④ $\sqrt{2}$ ⑤ $\sqrt{2x}$ ⑥ $\sqrt{x - 1}$ (3) ① $2\sqrt{3}\approx 3.46$ ② $2\sqrt{2}\approx 2.83$ ③ $\dfrac{3}{4}\sqrt{2}\approx 1.06$ ④ $\dfrac{\sqrt{6}}{2}\approx 1.22$
解析
(3)①$\frac{6}{\sqrt{3}}=\frac{6\sqrt{3}}{\sqrt{3}×\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3}\approx2×1.732\approx3.46$；②$\frac{2\sqrt{10}}{\sqrt{5}}=2\sqrt{\frac{10}{5}}=2\sqrt{2}\approx2×1.414\approx2.83$；③$\frac{\sqrt{63}}{2\sqrt{14}}=\frac{3\sqrt{7}}{2\sqrt{14}}=\frac{3\sqrt{7}}{2\sqrt{2}×\sqrt{7}}=\frac{3}{2\sqrt{2}}=\frac{3\sqrt{2}}{4}\approx\frac{3×1.414}{4}\approx1.06$；④$\sqrt{1.5}=\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2}\approx\frac{2.449}{2}\approx1.22$
