Bài 70 (Trang 40 SGK Toán 9, Tập 1):

Tìm giá trị các biểu thức sau bằng cách biến đổi, rút gọn thích hợp:

$\mathrm{a}) \sqrt{\frac{25}{81}. \frac{16}{49}. \frac{196}{9}};$

$\mathrm{b}) \sqrt{3\frac{1}{16}.2\frac{14}{25}.2\frac{34}{81}};$

$\mathrm{c}) \frac{\sqrt{640}.\sqrt{34,3}}{\sqrt{567}};$

$\mathrm{d}) \sqrt{21,6}.\sqrt{810}.\sqrt{{11}^2-5^2}.$

Hướng dẫn giải:

$$\begin{gathered} \mathrm{a}) \sqrt{\frac{25}{81}.\frac{16}{49}.\frac{196}{9}} = \sqrt{\frac{25}{81}}.\sqrt{\frac{16}{49}}.\sqrt{\frac{196}{9}} = \frac{\sqrt{25}}{\sqrt{81}}.\frac{\sqrt{16}}{49}.\frac{\sqrt{196}}{9} \\ = \frac{\sqrt{5^2}}{\sqrt{9^2}}.\frac{\sqrt{4^2}}{\sqrt{7^2}}.\frac{\sqrt{{14}^2}}{\sqrt{3^2}} = \frac{5.4.14}{9.7.3}= \frac{40}{27}. \end{gathered}$$

$$\begin{gathered} \mathrm{b}) \sqrt{3\frac{1}{16}.2\frac{14}{25}.2\frac{34}{81}} = \sqrt{\frac{49}{16}.\frac{64}{25}.\frac{196}{81}} = \sqrt{\frac{49}{16}}.\sqrt{\frac{64}{25}}.\sqrt{\frac{196}{81}} \\ =\frac{\sqrt{49}}{\sqrt{16}}.\frac{\sqrt{64}}{\sqrt{25}}.\frac{\sqrt{196}}{\sqrt{81}}=\frac{\sqrt{7^2}}{\sqrt{4^2}}.\frac{\sqrt{8^2}}{\sqrt{5^2}}.\frac{\sqrt{{14}^2}}{9^2} = \frac{7.8.14}{4.5.9} = \frac{196}{45}. \end{gathered}$$

$\mathrm{c}) \frac{\sqrt{640}.\sqrt{34,3}}{\sqrt{567}} = \frac{\sqrt{640.34,3}}{\sqrt{567}} = \sqrt{\frac{64.343}{567}} = \sqrt{\frac{64.49.7}{81.7}} = \sqrt{\frac{64.49}{81}} = \frac{8.7}{9} = \frac{56}{9}.$

$$\begin{gathered} \mathrm{d}) \sqrt{21,6}.\sqrt{810}.\sqrt{{11}^2-5^2} = \sqrt{21,6.810}.\sqrt{\left(11-5\right)\left(11+5\right)} =\sqrt{216.81}.\sqrt{6.16} \\ = \sqrt{216.6.81.16} = \sqrt{1296.81.16 } = \sqrt{{36}^2.9^2.4^2} = 36.9.4 = 1296. \end{gathered}$$