Bài 62 (Trang 33 SGK Toán 9, Tập 1):

Rút gọn các biểu thức sau:

a)$\frac{1}{2}\sqrt{48} - 2\sqrt{75} -\frac{\sqrt{33}}{\sqrt{11}} +5 \sqrt{1\frac{1}{3}}$ ;

b) $\sqrt{150} +\sqrt{1,6}.\sqrt{60}+ 4,5\sqrt{2\frac{2}{3}}-\sqrt{6}$;

c) $\left(\sqrt{28} - 2\sqrt{3} +\sqrt{7}\right)\sqrt{7} +\sqrt{84}$;

d) ${\left(\sqrt{6} +\sqrt{5}\right)}^{2 } - \sqrt{120}$.

Hướng dẫn giải:

$$\begin{gathered} \mathrm{a}) \frac{1}{2}\sqrt{48 }- 2\sqrt{75} - \frac{\sqrt{33}}{\sqrt{11}} + 5\sqrt{1\frac{1}{3}} = \frac{1}{2}\sqrt{16.3 } - 2\sqrt{75} -\frac{\sqrt{33}}{\sqrt{11}} + 5\sqrt{\frac{4.4}{3.3}} \\ = \frac{4}{2}\sqrt{3} - 2.5\sqrt{3} +\frac{5.2}{3}\sqrt{3} = \frac{4}{2}\sqrt{3} - 2.5\sqrt{3} +\frac{5.2}{3}\sqrt{3} = 2\sqrt{3 }- 10\sqrt{3} -\sqrt{3}+ \frac{10}{3}\sqrt{3} \\ = -9\sqrt{3} +\frac{10}{3}\sqrt{3} = \left(-9 +\frac{10}{3}\right)\sqrt{3} = -\frac{17}{3}\sqrt{3} \end{gathered}$$

$$\begin{gathered} \mathrm{b}) \sqrt{150} + \sqrt{1,6}.\sqrt{60} 4,5\sqrt{2\frac{2}{3}}-\sqrt{6 }= \sqrt{25.6} +\sqrt{1,6}.\sqrt{60} + 4,5\sqrt{2\frac{2}{3}} -\sqrt{6} \\ = 5\sqrt{6 } +\sqrt{16.6} +4,5.\frac{1}{3}\sqrt{3^2 \frac{4.2}{3}} -\sqrt{6} = 5\sqrt{6} +4\sqrt{6} +4,5.\frac{2}{3}\sqrt{6}-\sqrt{6} \\ = 9\sqrt{6} +3\sqrt{6} -\sqrt{6} = 11\sqrt{6} \end{gathered}$$

$$\begin{gathered} \mathrm{c}) \left(\sqrt{28 } -2\sqrt{3} +\sqrt{7}\right)\sqrt{7} +\sqrt{84} = \left(\sqrt{4.7} -2\sqrt{3} +\sqrt{7}\right)\sqrt{7} +\sqrt{4.21} \\ = \left(2\sqrt{7 }-2\sqrt{3} +\sqrt{7}\right)\sqrt{7 }+2\sqrt{21 }= \left(3\sqrt{7}- 2\sqrt{7}\right)\sqrt{7} +2\sqrt{7} \\ = 3.7 - 2\sqrt{21} +2\sqrt{21} = 21. \end{gathered}$$

$$\begin{gathered} \mathrm{d}) {\left(\sqrt{6} +\sqrt{5}\right)}^2 -\sqrt{120} = {\left(\sqrt{6}\right)}^2 +2\sqrt{6}.\sqrt{5} +{\left(\sqrt{5}\right)}^2 -\sqrt{4.30} \\ = 6 + 2\sqrt{30} +5 -2\sqrt{30} = 11 \end{gathered}$$