Упр.17.28 ГДЗ Мерзляк Поляков 8 класс (Алгебра)

1. $$\frac{\sqrt{y}+4}{\sqrt{xy}+y}-\frac{\sqrt{x}-4}{x+\sqrt{xy}}$$
   $$=\frac{\sqrt{y}+4}{\sqrt{y}(\sqrt{x}+\sqrt{y})}-\frac{\sqrt{x}-4}{\sqrt{x}(\sqrt{x}+\sqrt{y})}$$
   $$=\frac{\sqrt{x}(\sqrt{y}+4)-\sqrt{y}(\sqrt{x}-4)}{\sqrt{xy}(\sqrt{x}+\sqrt{y})}$$
   $$=\frac{\sqrt{xy}+4\sqrt{x}-\sqrt{xy}+4\sqrt{y}}{\sqrt{xy}(\sqrt{x}+\sqrt{y})}$$
   $$=\frac{4(\sqrt{x}+\sqrt{y})}{\sqrt{xy}(\sqrt{x}+\sqrt{y})}=\frac{4}{\sqrt{xy}}.$$

2. $$\frac{\sqrt{a}}{\sqrt{a}+4}-\frac{a}{a-16}$$
   $$=\frac{\sqrt{a}}{\sqrt{a}+4}-\frac{a}{(\sqrt{a}-4)(\sqrt{a}+4)}$$
   $$=\frac{\sqrt{a}(\sqrt{a}-4)-a}{(\sqrt{a}-4)(\sqrt{a}+4)}$$
   $$=\frac{a-4\sqrt{a}-a}{a-16}=\frac{-4\sqrt{a}}{a-16}=\frac{4\sqrt{a}}{16-a}.$$

3. $$\frac{a}{\sqrt{ab}-b}+\frac{\sqrt{b}}{\sqrt{b}-\sqrt{a}}$$
   $$=\frac{a}{\sqrt{b}(\sqrt{a}-\sqrt{b})}-\frac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}$$
   $$=\frac{a-\sqrt{b}\cdot\sqrt{b}}{\sqrt{b}(\sqrt{a}-\sqrt{b})}$$
   $$=\frac{a-b}{\sqrt{b}(\sqrt{a}-\sqrt{b})}$$
   $$=\frac{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}{\sqrt{b}(\sqrt{a}-\sqrt{b})}=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{b}}.$$

4. $$\frac{a+\sqrt{a}}{\sqrt{b}}\cdot\frac{b}{2\sqrt{a}+2}$$
   $$=\frac{\sqrt{a}(\sqrt{a}+1)}{\sqrt{b}}\cdot\frac{b}{2(\sqrt{a}+1)}$$
   $$=\frac{\sqrt{a}\cdot b}{2\sqrt{b}}=\frac{\sqrt{ab}\cdot\sqrt{b}}{2}=\frac{\sqrt{ab}}{2}.$$

5. $$\frac{\sqrt{c}-5}{\sqrt{c}}:\frac{c-25}{3c}$$
   $$=\frac{\sqrt{c}-5}{\sqrt{c}}\cdot\frac{3c}{c-25}$$
   $$=\frac{(\sqrt{c}-5)\cdot 3c}{\sqrt{c}(\sqrt{c}-5)(\sqrt{c}+5)}$$
   $$=\frac{3\sqrt{c}}{\sqrt{c}+5}.$$

6. $$\left(\sqrt{a}-\frac{a}{\sqrt{a}+1}\right):\frac{\sqrt{a}}{a-1}$$
   $$=\frac{\sqrt{a}(\sqrt{a}+1)-a}{\sqrt{a}+1}\cdot\frac{a-1}{\sqrt{a}}$$
   $$=\frac{a+\sqrt{a}-a}{\sqrt{a}+1}\cdot\frac{a-1}{\sqrt{a}}$$
   $$=\frac{\sqrt{a}}{\sqrt{a}+1}\cdot\frac{(\sqrt{a}-1)(\sqrt{a}+1)}{\sqrt{a}}=\sqrt{a}-1.$$

7. $$\left(\frac{\sqrt{a}+\sqrt{b}}{\sqrt{b}}+\frac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right):\frac{\sqrt{a}}{\sqrt{b}}$$
   $$=\frac{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})+\sqrt{b}\cdot\sqrt{b}}{\sqrt{b}(\sqrt{a}-\sqrt{b})}\cdot\frac{\sqrt{b}}{\sqrt{a}}$$
   $$=\frac{a-b+b}{\sqrt{b}(\sqrt{a}-\sqrt{b})}\cdot\frac{\sqrt{b}}{\sqrt{a}}$$
   $$=\frac{a}{(\sqrt{a}-\sqrt{b})\sqrt{a}}=\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}.$$

8. $$\left(\frac{\sqrt{x}-3}{\sqrt{x}+3}+\frac{12\sqrt{x}}{x-9}\right):\frac{\sqrt{x}+3}{x-3\sqrt{x}}$$
   $$=\left(\frac{\sqrt{x}-3}{\sqrt{x}+3}+\frac{12\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}\right)\cdot\frac{x-3\sqrt{x}}{\sqrt{x}+3}$$
   $$=\frac{(\sqrt{x}-3)^2+12\sqrt{x}}{(\sqrt{x}+3)(\sqrt{x}-3)}\cdot\frac{x-3\sqrt{x}}{\sqrt{x}+3}$$
   $$=\frac{x-6\sqrt{x}+9+12\sqrt{x}}{(\sqrt{x}+3)(\sqrt{x}-3)}\cdot\frac{\sqrt{x}(\sqrt{x}-3)}{\sqrt{x}+3}$$
   $$=\frac{(\sqrt{x}+3)^2\cdot\sqrt{x}}{(\sqrt{x}+3)^2}=\sqrt{x}.$$

Ответ

1) $$\frac{4}{\sqrt{xy}}$$; 2) $$\frac{4\sqrt{a}}{16-a}$$; 3) $$\frac{\sqrt{a}+\sqrt{b}}{\sqrt{b}}$$; 4) $$\frac{\sqrt{ab}}{2}$$; 5) $$\frac{3\sqrt{c}}{\sqrt{c}+5}$$; 6) $$\sqrt{a}-1$$; 7) $$\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}$$; 8) $$\sqrt{x}$$.