# Bài 50 trang 30 SGK Toán 9 tập 1

##### Hướng dẫn giải

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

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

+ $$\sqrt{A^2 B}=A\sqrt{B}$$,   nếu $$A,\ B \ge 0$$.

+ $$\sqrt{A^2 B}=-A\sqrt B$$,   nếu $$A < 0,\ B \ge 0$$.

Lời giải chi tiết

+ Ta có:

$$\dfrac{5}{\sqrt{10}}=\dfrac{5.\sqrt{10}}{\sqrt{10}.\sqrt{10}}=\dfrac{5\sqrt{10}}{(\sqrt{10})^2}=\dfrac{5\sqrt{10}}{10}=\dfrac{5.\sqrt{10}}{5.2}$$

$$=\dfrac{\sqrt{10}}{2}$$.

+ Ta có:

$$\dfrac{5}{2\sqrt{5}}=\dfrac{5.\sqrt 5}{2\sqrt 5.\sqrt 5}=\dfrac{5\sqrt{5}}{2.(\sqrt 5.\sqrt 5)}=\dfrac{5\sqrt{5}}{2(\sqrt 5)^2}$$

$$=\dfrac{5\sqrt 5}{2.5}=\dfrac{\sqrt 5}{2}$$.

+ Ta có:

$$\dfrac{1}{3\sqrt{20}}=\dfrac{1.\sqrt{20}}{3\sqrt{20}.\sqrt{20}}=\dfrac{\sqrt{20}}{3.(\sqrt{20}.\sqrt{20})}=\dfrac{\sqrt{20}}{3.(\sqrt{20})^2}$$

$$=\dfrac{\sqrt{20}}{3.20}=\dfrac{\sqrt{2^2.5}}{60}=\dfrac{2\sqrt 5}{60}=\dfrac{2\sqrt 5}{2.30}=\dfrac{\sqrt 5}{30}$$.

+ Ta có:

$$\dfrac{(2\sqrt{2}+2)}{5.\sqrt 2}=\dfrac{(2\sqrt 2+2).\sqrt 2}{5\sqrt 2. \sqrt 2}=\dfrac{2\sqrt 2.\sqrt 2+2.\sqrt 2}{5.(\sqrt 2)^2}$$

$$=\dfrac{2.2+2\sqrt 2}{5.2}=\dfrac{2(2+\sqrt 2)}{5.2}=\dfrac{2+\sqrt 2}{5}$$.

+ Ta có:

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

$$= \dfrac{y\sqrt y+b(\sqrt y)^2}{by}=\dfrac{y\sqrt y+by}{by}$$

$$=\dfrac{y(\sqrt y+b)}{b.y}=\dfrac{\sqrt y+b}{b}$$.