# Bài 30 trang 19 SGK Toán 9 tập 1

##### Hướng dẫn giải

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

+) $$\sqrt{a^2}=|a|$$.

+) $$|a| =a$$,  nếu $$a \ge 0$$.

$$|a|=-a$$,  nếu $$a <0$$.

+) $$a^{m.n}=a^m.a^n$$,   với $$m,\ n \in \mathbb{N}$$.

Lời giải chi tiết

a) Ta có:

$$\dfrac{y}{x}.\sqrt{\dfrac{x^{2}}{y^{4}}}=\dfrac{y}{x}.\dfrac{\sqrt{x^2}}{\sqrt{y^{4}}}$$

$$=\dfrac{y}{x}.\dfrac{\sqrt{x^2}}{\sqrt{(y^2)^2}}=\dfrac{y}{x}.\dfrac{|x|}{|y^2|}$$

Vì $$x> 0$$ nên $$|x|=x$$.

Vì $$y \ne 0$$  nên  $$y^2 > 0 \Rightarrow |y^2|=y^2$$.

$$\Rightarrow \dfrac{y}{x}.\dfrac{|x|}{|y^2|} =\dfrac{y}{x}.\dfrac{x}{y^2}=\dfrac{y}{x}.\dfrac{x}{y.y}=\dfrac{1}{y}$$.

Vậy $$\dfrac{y}{x}.\sqrt{\dfrac{x^{2}}{y^{4}}}=\dfrac{1}{y}$$.

b) Ta có:

$$2y^2.\sqrt{\dfrac{x^{4}}{4y^{2}}}=2y^2.\dfrac{\sqrt{x^4}}{\sqrt{4y^2}}=2y^2.\dfrac{\sqrt{(x^2)^2}}{\sqrt{2^2.y^2}}$$

$$=2y^2.\dfrac{\sqrt{(x^2)^2}}{\sqrt{(2y)^2}}=2y^2.\dfrac{|x^2|}{|2y|}$$

Vì $$x^2 \ge 0 \Rightarrow |x^2|=x^2$$.

Vì $$y<0$$  nên  $$2y < 0 \Rightarrow |2y|=-2y$$

$$\Rightarrow 2y^2.\dfrac{|x^2|}{|2y|}=2y^2.\dfrac{x^2}{-2y}=\dfrac{2y^2.x^2}{-2y}$$

$$=\dfrac{x^2.y.2y}{-2y}=-x^2y$$.

Vậy $$2y^2.\sqrt{\dfrac{x^{4}}{4y^{2}}}=-x^2y$$.

c) Ta có:

$$5xy.\sqrt{\dfrac{25x^{2}}{y^{6}}}=5xy.\dfrac{\sqrt{25x^2}}{\sqrt{y^6}}=\dfrac{\sqrt{5^2.x^2}}{\sqrt{(y^3)^2}}$$

$$=\dfrac{\sqrt{(5x)^2}}{\sqrt{(y^3)^2}}=5xy.\dfrac{|5x|}{|y^3|}$$

Vì $$x<0 \Rightarrow |5x| = - 5x$$

Vì $$y>0 \Rightarrow y^3 >0 \Rightarrow |y^3|=y^3$$.

$$\Rightarrow 5xy.\dfrac{|5x|}{|y^3|}=5xy.\dfrac{-5x}{y^3}=\dfrac{5xy.(-5x)}{y^3}$$

$$=\dfrac{[5.(-5)].(x.x).y}{y^2.y}=\dfrac{-25x^2}{y^2}$$

Vậy $$5xy.\sqrt{\dfrac{25x^{2}}{y^{6}}}=\dfrac{-25x^2}{y^2}$$.

d) Ta có:

$$0,2x^{3}y^{3}.\sqrt{\dfrac{16}{x^{4}y^{8}}}=0,2x^3y^3.\dfrac{\sqrt{16}}{\sqrt{x^4y^8}}$$

$$=0,2x^3y^3\dfrac{\sqrt{4^2}}{\sqrt{(x^2)^2.(y^4)^2}}$$

$$=0,2x^3y^3.\dfrac{\sqrt{4^2}}{\sqrt{(x^2)^2}.\sqrt{(y^4)^2}}=0,2x^3y^3.\dfrac{4}{|x^2|.|y^4|}$$.

Vì $$x \ne 0,\ y \ne 0$$  nên  $$x^2 > 0$$  và $$y^4 > 0$$

$$\Rightarrow |x^2| =x^2$$  và $$|y^4|=y^4$$.

$$\Rightarrow 0,2x^3y^3.\dfrac{4}{|x^2|.|y^4|}=0,2x^3y^3.\dfrac{4}{x^2y^4}$$

$$=\dfrac{0,2x^3y^3.4}{x^2y^4}$$

$$=\dfrac{(0,2.4).(x^2.x).y^3}{x^2.(y^3.y)}=\dfrac{0,8.x.x^2y^3}{y.x^2y^3}=\dfrac{0,8x}{y}$$.

Vậy $$0,2x^{3}y^{3}.\sqrt{\dfrac{16}{x^{4}y^{8}}}=\dfrac{0,8x}{y}$$.