Bài 47 (Trang 27 SGK Toán 9, Tập 1):

Rút gọn:

$$\begin{gathered} \mathrm{a}) \frac{2}{{\mathrm{x}}^2-{\mathrm{y}}^2}.\sqrt{\frac{3{\left(\mathrm{x}+\mathrm{y}\right)}^2}{2} } \mathrm{với} \mathrm{x} \ge 0, \mathrm{y} \ge 0, \mathrm{x} \ne \mathrm{y}; \\ \mathrm{b}) \frac{2}{2\mathrm{a}-1}.\sqrt{5{\mathrm{a}}^2\left(1-4\mathrm{a}+4{\mathrm{a}}^2\right)} \mathrm{với} \mathrm{a} > 0,5. \end{gathered}$$

Hướng dẫn giải:

$\mathrm{a}) \frac{2}{{\mathrm{x}}^2-{\mathrm{y}}^2}.\sqrt{\frac{3{\left(\mathrm{x}+\mathrm{y}\right)}^2}{2}} = \frac{2}{{\mathrm{x}}^2-{\mathrm{y}}^2}\left|\mathrm{x}+\mathrm{y}\right|\sqrt{\frac{3}{2}} = \frac{\mathrm{x} + \mathrm{y}}{{\mathrm{x}}^2 - {\mathrm{y}}^2}\sqrt{2^2.\frac{3}{2}} = \frac{\sqrt{6}}{\mathrm{x}-\mathrm{y}} \left(\mathrm{vì} \mathrm{x} \ge 0, \mathrm{y} \ge 0, \mathrm{x} \ne \mathrm{y} \mathrm{nên} \mathrm{x} + \mathrm{y} > 0\right).$

$$\begin{gathered} \mathrm{b}) \frac{2}{2\mathrm{a}-1}\sqrt{5{\mathrm{a}}^2\left(1-4\mathrm{a}+4{\mathrm{a}}^2\right)} = \frac{2}{2\mathrm{a}-1}\sqrt{5{\mathrm{a}}^2{\left(1-2\mathrm{a}\right)}^2} =\frac{2\left|\mathrm{a}\right|.\left|1-2\mathrm{a}\right|\sqrt{5}}{2\mathrm{a}-1} = \frac{2.\mathrm{a}\left(2\mathrm{a}-1\right)\sqrt{5}}{2\mathrm{a}-1} = 2\sqrt{5}\mathrm{a} \\ \left(\mathrm{Vì} \mathrm{a} > 0,5 \mathrm{nên} \mathrm{a} > 0 \Rightarrow 1 - 2\mathrm{a} < 0\right) \end{gathered}$$