Bài 19 (Trang 15 SGK Toán 9, Tập 1):

Rút gọn các biểu thức sau:

 $$\begin{gathered} \mathrm{a}) \sqrt{0,36{\mathrm{a}}^2} \mathrm{với} \mathrm{a}<0; \\ \mathrm{b}) \sqrt{{\mathrm{a}}^4{\left(3-\mathrm{a}\right)}^2} \mathrm{với} \mathrm{a}\ge 3; \\ \mathrm{c}) \sqrt{27.48{\left(1-\mathrm{a}\right)}^2} \mathrm{với} \mathrm{a}>1; \\ \mathrm{d}) \frac{1}{\mathrm{a}-\mathrm{b}}\sqrt{{\mathrm{a}}^4{\left(\mathrm{a}-\mathrm{b}\right)}^2}\mathrm{với} \mathrm{a}>\mathrm{b} \end{gathered}$$

Hướng dẫn giải:

$\mathrm{a}) \sqrt{0,36{\mathrm{a}}^2} =\sqrt{0,36}.\sqrt{{\mathrm{a}}^2} = 0,6.\left|\mathrm{a}\right| = -0,6\mathrm{a} (\mathrm{vì} \mathrm{a}<0)$

$b) \sqrt{a^4{\left(3-a\right)}^2} = \sqrt{a^4}.\sqrt{{\left(3-a\right)}^2} = \left|a^2\right|.\left|3-a\right| = a^2\left(a-3\right) (vì a\ge 3 nên a-3 \ge 0)$

$\mathrm{c}. \sqrt{27.48{\left(1-\mathrm{a}\right)}^2} = \sqrt{9^2.4^2{\left(1-\mathrm{a}\right)}^2} = \sqrt{9^2}.\sqrt{4^2}.\sqrt{{\left(1-\mathrm{a}\right)}^2} = 9.4.\left|1-\mathrm{a}\right| = 36.\left(1-\mathrm{a}\right) (\mathrm{vì} \mathrm{a}>1 \mathrm{nên} 1-\mathrm{a}<0)$

$$\begin{gathered} \mathrm{d}) \frac{1}{\mathrm{a}-\mathrm{b}}\sqrt{{\mathrm{a}}^4{\left(\mathrm{a}-\mathrm{b}\right)}^2} = \frac{1}{\mathrm{a}-\mathrm{b}}.\sqrt{{\mathrm{a}}^4}.\sqrt{{\left(\mathrm{a}-\mathrm{b}\right)}^2} = \frac{1}{\mathrm{a}-\mathrm{b}}.\left|{\mathrm{a}}^2\right|.\left|\mathrm{a}-\mathrm{b}\right| \\ = \frac{1}{\mathrm{a}-\mathrm{b}}.{\mathrm{a}}^2.(\mathrm{a}-\mathrm{b}) = {\mathrm{a}}^2 ( \mathrm{vì} {\mathrm{a}}^2\ge 0\Rightarrow \mathrm{a}>\mathrm{b}\iff \mathrm{a}-\mathrm{b}>0) \end{gathered}$$