Bài 34 (Trang 19 SGK Toán 9, Tập 1):

Rút gọn các biểu thức sau:

$\mathrm{a}) {\mathrm{ab}}^2.\sqrt{\frac{3}{{\mathrm{a}}^2{\mathrm{b}}^4}} \mathrm{với} \mathrm{a} < 0, \mathrm{b}\ne 0;$

$\mathrm{b}) \sqrt{\frac{27{\left(\mathrm{a}-3\right)}^2}{48}} \mathrm{với} \mathrm{a} > 3;$

$\mathrm{c}) \sqrt{\frac{9 + 12\mathrm{a} + 4{\mathrm{a}}^2}{{\mathrm{b}}^2}} \mathrm{với} \mathrm{a} \ge -1,5 \mathrm{và} \mathrm{b} < 0;$

$\mathrm{d}) \left(\mathrm{a} - \mathrm{b}\right).\sqrt{\frac{\mathrm{ab}}{{\left(\mathrm{a} - \mathrm{b}\right)}^2}} \mathrm{với} \mathrm{a} < \mathrm{b} < 0.$

Hướng dẫn giải:

$\mathrm{a}) {\mathrm{ab}}^2.\sqrt{\frac{3}{{\mathrm{a}}^2{\mathrm{b}}^4}} \mathrm{với} \mathrm{a} < 0, \mathrm{b}\ne 0$

Ta có:

$$\begin{gathered} {\mathrm{ab}}^2.\sqrt{\frac{3}{{\mathrm{a}}^2{\mathrm{b}}^4}} = a{b}^2.\frac{\sqrt{3}}{\sqrt{a^2{b}^4}} = a{b}^2.\frac{\sqrt{3}}{\sqrt{a^2.\sqrt{b^4}}} \\ = a{b}^2.\frac{\sqrt{3}}{\left|a\right|.\left|b^2\right|} = a{b}^2.\frac{\sqrt{3}}{-a{b}^2} = -\sqrt{3} (Vì a < 0, b\ne 0 nên b^2 > 0 ) \Rightarrow \left|b^2\right| = b^2) \end{gathered}$$

$\mathrm{b}) \sqrt{\frac{27{\left(\mathrm{a}-3\right)}^2}{48}} \mathrm{với} \mathrm{a} > 3$

Ta có:

$$\begin{gathered} \sqrt{\frac{27{\left(\mathrm{a}-3\right)}^2}{48}} = \sqrt{\frac{27}{48}.{\left(\mathrm{a} - 3\right)}^2} = \sqrt{\frac{27}{48}}.\sqrt{{\left(\mathrm{a} - 3\right)}^2} \\ = \sqrt{\frac{9.3}{16.3}}.\sqrt{{\left(\mathrm{a} - 3\right)}^2} = \sqrt{\frac{9}{16}}.\sqrt{{\left(\mathrm{a} -3\right)}^2} \\ = \frac{3}{4}.\sqrt{{\left(\mathrm{a}-3\right)}^2} = \frac{3}{4}.\left|\mathrm{a}-3\right| = \frac{3}{4}\left(\mathrm{a} - 3\right) \\ \mathit{(}Vì\mathit{ }a\mathit{ }\mathit{>}\mathit{ }\mathit{3}\mathit{ }nên\mathit{ }a\mathit{ }\mathit{-}\mathit{ }\mathit{3}\mathit{ }\mathit{>}\mathit{ }\mathit{0}\mathit{ }\mathit{\Rightarrow }\mathit{|}\mathit{a}\mathit{ }\mathit{-}\mathit{ }\mathit{3}\mathit{|}\mathit{ }\mathit{=}\mathit{ }a\mathit{ }\mathit{-}\mathit{ }\mathit{3}\mathit{)} \end{gathered}$$

$\mathrm{c}) \sqrt{\frac{9 + 12\mathrm{a} + 4{\mathrm{a}}^2}{{\mathrm{b}}^2}} \mathrm{với} \mathrm{a} \ge -1,5 \mathrm{và} \mathrm{b} < 0;$

Ta có:

$$\begin{gathered} \sqrt{\frac{9 + 12\mathrm{a} + 4{\mathrm{a}}^2}{{\mathrm{b}}^2}} = \sqrt{\frac{3^2 +\hspace{0.17em}2.3.2a + 2^2.a^2}{b^2}} \\ = \sqrt{\frac{3^2 + 2.3.2a + {\left(2a\right)}^2}{b^2}} = \sqrt{\frac{{\left(3 +\hspace{0.17em}2a\right)}^2}{b^2}} = \frac{\left|3 + 2a\right|}{\left|b\right|} \end{gathered}$$

$$\begin{gathered} \mathrm{Vì} \mathrm{a} \ge -1,5 \Rightarrow \mathrm{a} +\hspace{0.17em}1,5 > 0 \\ \iff 2(\mathrm{a} + 1,5) > 0 \\ \iff 2\mathrm{a} \hspace{0.17em}+3 > 0 \\ \iff 3 + 2\mathrm{a} > 0 \\ \iff \left|3 + 2\mathrm{a}\right| = 3 +\hspace{0.17em}2\mathrm{a} \end{gathered}$$

$\mathrm{Vì} \mathrm{b} < 0 \Rightarrow \left|\mathrm{b}\right| = -\mathrm{b}$

Do vậy, biểu thức $\frac{\left|3 + 2\mathrm{a}\right|}{\left|\mathrm{b}\right|} = \frac{3 + 2\mathrm{a}}{-\mathrm{b}} =-\frac{3 + 2\mathrm{a}}{\mathrm{b}}$.

Vậy $\sqrt{\frac{9 + 12\mathrm{a} + 4{\mathrm{a}}^2}{{\mathrm{b}}^2}} = -\frac{3 + 2\mathrm{a}}{\mathrm{b}}$.

$\mathrm{d}) \left(\mathrm{a} - \mathrm{b}\right).\sqrt{\frac{\mathrm{ab}}{{\left(\mathrm{a} - \mathrm{b}\right)}^2}} \mathrm{với} \mathrm{a} < \mathrm{b} < 0.$

Ta có:

$\left(\mathrm{a} - \mathrm{b}\right).\sqrt{\frac{\mathrm{ab}}{{\left(\mathrm{a} - \mathrm{b}\right)}^2}} = \left(a - b\right).\frac{\sqrt{ab}}{\sqrt{{\left(a-b\right)}^2}} = \left(a-b\right)\frac{\sqrt{ab}}{\left|a-b\right|} = \left(a-b\right)\frac{\sqrt{ab}}{-\left(a-b\right)} = -\sqrt{ab}$

$(Vì a < b < 0 nên a - b < 0 \Rightarrow \left|a - b\right| = -\left(a - b\right) và a.b > 0)$