# Bài 32 trang 19 SGK Toán 9 tập 1

##### Hướng dẫn giải

+ Sử dụng công thức đổi hỗn số ra phân số:

$$a\dfrac{b}{c}=\dfrac{a.b+c}{b}$$.

+ $$\sqrt{a^2}=a$$ ,  với $$a \ge 0$$.

+ $$\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}},$$   với  $$a \ge 0,\ b>0$$.

+ $$\sqrt{ab}=\sqrt{a}. \sqrt{b}$$,   với $$a,\ b \ge 0$$.

+ $$a^2 -b^2=(a-b)(a+b)$$

Lời giải chi tiết

a) Ta có:

$$\sqrt{1\dfrac{9}{16}.5\dfrac{4}{9}.0,01}=\sqrt{\dfrac{1.16+9}{16}.\dfrac{5.9+4}{9}.\dfrac{1}{100}}$$

$$=\sqrt{\dfrac{16+9}{16}.\dfrac{45+4}{9}.\dfrac{1}{100}}$$

$$=\sqrt{\dfrac{25}{16}.\dfrac{49}{9}.\dfrac{1}{100}}$$

$$=\sqrt{\dfrac{25}{16}}.\sqrt{\dfrac{49}{9}}.\sqrt{\dfrac{1}{100}}$$

$$=\dfrac{\sqrt{25}}{\sqrt{16}}.\dfrac{\sqrt{49}}{\sqrt{9}}.\dfrac{\sqrt{1}}{\sqrt{100}}$$

$$=\dfrac{\sqrt{5^2}}{\sqrt{4^2}}.\dfrac{\sqrt{7^2}}{\sqrt{3^2}}.\dfrac{1}{\sqrt{10^2}}$$

$$=\dfrac{5}{4}.\dfrac{7}{3}.\dfrac{1}{10}=\dfrac{5.7.1}{4.3.10}=\dfrac{35}{120}=\dfrac{7}{24}.$$

b) Ta có:

$$\sqrt{1,44.1,21-1,44.0,4} = \sqrt{1,44(1,21-0,4)}$$

$$=\sqrt{1,44.0,81}$$

$$=\sqrt{1,44}.\sqrt{0,81}$$

$$=\sqrt{1,2^2}.\sqrt{0,9^2}$$

$$=1,2.0,9=1,08$$.

c) Ta có:

$$\sqrt{\dfrac{165^{2}-124^{2}}{164}}$$$$=\sqrt{\dfrac{(165-124)(165+124)}{164}}$$

$$=\sqrt{\dfrac{41.289}{41.4}}$$ $$=\sqrt{\dfrac{289}{4}}$$

$$=\dfrac{\sqrt{289}}{\sqrt{4}}$$ $$=\dfrac{\sqrt{17^2}}{\sqrt{2^2}}$$ $$=\dfrac{17}{2}$$.

Câu d: Ta có:

$$\sqrt{\dfrac{149^{2}-76^{2}}{457^{2}-384^{2}}}$$ $$=\sqrt{\dfrac{(149-76)(149+76)}{(457-384)(457+384)}}$$

$$=\sqrt{\dfrac{73.225}{73.841}}$$ $$=\sqrt{\dfrac{225}{841}}$$

$$=\dfrac{15^2}{29^2}=\dfrac{15}{29}$$.