# Bài 54 trang 30 SGK Toán 9 tập 1

##### Hướng dẫn giải

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

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

Lời giải chi tiết

+ Ta có:

$$\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{(\sqrt 2)^2+ \sqrt 2}{1+ \sqrt 2}=\dfrac{\sqrt{2}(\sqrt{2}+1)}{1+\sqrt{2}}$$

$$=\dfrac{\sqrt 2(1+ \sqrt 2)}{\sqrt 2}=\sqrt{2}$$.

+ Ta có:

$$\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{3.5}-\sqrt{5.1}}{1-\sqrt{3}}$$$$=\dfrac{\sqrt{5}.\sqrt{3}-\sqrt{5}.1}{1-\sqrt{3}}$$

$$=\dfrac{\sqrt{5}(\sqrt{3}-1)}{1-\sqrt{3}}=\dfrac{-\sqrt{5}(1-\sqrt{3})}{1-\sqrt{3}}=-\sqrt{5}$$.

+ Ta có:

$$\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{(\sqrt 2)^2.\sqrt 3-\sqrt 6}{\sqrt{4.2}- 2}$$

$$=\dfrac{\sqrt 2.(\sqrt 2.\sqrt 3)-\sqrt 6}{2\sqrt 2 -2}$$$$=\dfrac{2\sqrt{6}-\sqrt 6}{2(\sqrt{2}-1)}$$

$$=\dfrac{\sqrt{6}(\sqrt{2}-1)}{2(\sqrt{2}-1)}=\dfrac{\sqrt{6}}{2}$$.

+ Ta có:

$$\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{(\sqrt a)^2-\sqrt a .1}{1-\sqrt a}=\dfrac{\sqrt{a}(\sqrt{a}-1)}{1-\sqrt{a}}$$

$$=\dfrac{-\sqrt{a}(1-\sqrt{a})}{1-\sqrt{a}}=-\sqrt{a}$$.

+ Ta có:

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