# Bài 63 trang 33 SGK Toán 9 tập 1

##### Hướng dẫn giải

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

+ $$\dfrac{A}{\sqrt B}=\dfrac{A\sqrt B}{B}$$,   với $$B > 0$$.

+ $$(\sqrt b)^2=b$$,  với $$b \ge 0$$.

Lời giải chi tiết

a) Ta có:

$$\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{b}{a}}$$

$$=\dfrac{\sqrt{a}}{\sqrt b}+\sqrt{ab}+\dfrac{a}{b}.\dfrac{\sqrt{b}}{\sqrt a}$$

$$=\dfrac{\sqrt{a}.\sqrt b}{(\sqrt b)^2}+\sqrt{ab}+\dfrac{a}{b}.\dfrac{\sqrt{b}.\sqrt a}{(\sqrt a)^2}$$

$$=\dfrac{\sqrt{ab}}{b}+\sqrt{ab}+\dfrac{a}{b}.\dfrac{\sqrt{ab}}{a}$$

$$=\dfrac{\sqrt{ab}}{b}+\sqrt{ab}+\dfrac{\sqrt{ab}}{b}$$

$$={\left(\dfrac{\sqrt{ab}}{b}+\dfrac{\sqrt{ab}}{b} \right)}+\sqrt{ab}$$

$$=\dfrac{2\sqrt{ab}}{b}+\sqrt{ab}$$

$$=\dfrac{2\sqrt{ab}}{b}+\dfrac{b\sqrt{ab}}{b}$$

$$=\dfrac{2+b}{b}\sqrt{ab}$$.

b) Ta có:

$$\sqrt{\dfrac{m}{1-2x+x^{2}}}.\sqrt{\dfrac{4m-8mx+4mx^{2}}{81}}$$

$$=\sqrt{\dfrac{m}{1-2x+x^{2}}}.\sqrt{\dfrac{4m(1-2x+x^{2})}{81}}$$

$$=\sqrt{\dfrac{m}{1-2x+x^{2}}.\dfrac{4m(1-2x+x^{2})}{81}}$$

$$=\sqrt{\dfrac{m}{1}.\dfrac{4m}{81}}=\sqrt{\dfrac{4m^{2}}{81}}$$

$$=\sqrt{\dfrac{(2m)^2}{9^2}}=\dfrac{|2m|}{9}=\dfrac{2m}{9}$$.

Vì $$m >0$$ nên $$|2m|=2m$$.