Задание 15 Параграф 17 ГДЗ Рабочая тетрадь 2 Мерзляк Полонский 8 класс (Алгебра)

1) (va+3)/(va-3)-(va+6)/va;
2) b/(b-25)-vb/(vb+5);
3) (4vm-vn)/(m+vmn)•(2m-2n)/(4m-vmn);
4) (vc-4)/vc :(c-16)/2c;
5) (va/(va-vb)+va/vb) :vb/(vb-va);
6) ((vc+7)/(vc-7)-(vc-7)/(vc+7)) :14/(c-7vc);
7) (va-6)/(a+3va)-(va-3)/va+va/(va+3);
8) (vm+4)/(m-6vm+9) :(m-16)/(2vm-6)-2/(vm-4).

1. $$\frac{\sqrt a+3}{\sqrt a-3}-\frac{\sqrt a+6}{\sqrt a}= \frac{\sqrt a(\sqrt a+3)-(\sqrt a+6)(\sqrt a-3)}{\sqrt a(\sqrt a-3)}$$

   $$=\frac{a+3\sqrt a-\left(a-3\sqrt a+6\sqrt a-18\right)}{\sqrt a(\sqrt a-3)} =\frac{18}{\sqrt a(\sqrt a-3)}=\frac{18}{a-3\sqrt a}.$$

2. $$\frac{b}{b-25}-\frac{\sqrt b}{\sqrt b+5} =\frac{b}{(\sqrt b-5)(\sqrt b+5)}-\frac{\sqrt b}{\sqrt b+5}$$

   $$=\frac{b-\sqrt b(\sqrt b-5)}{(\sqrt b-5)(\sqrt b+5)} =\frac{b-b+5\sqrt b}{b-25} =\frac{5\sqrt b}{b-25}.$$

3. $$\frac{4\sqrt m-\sqrt n}{m+\sqrt{mn}}\cdot\frac{2m-2n}{4m-\sqrt{mn}}$$

   $$=\frac{(4\sqrt m-\sqrt n)\cdot 2(m-n)}{\sqrt m(\sqrt m+\sqrt n)\cdot \sqrt m(4\sqrt m-\sqrt n)}$$

   $$=\frac{2(\sqrt m-\sqrt n)(\sqrt m+\sqrt n)\cdot(\sqrt m-\sqrt n)}{m(\sqrt m+\sqrt n)} =\frac{2(\sqrt m-\sqrt n)}{m}.$$

4. $$\frac{\sqrt c-4}{\sqrt c}:\frac{c-16}{2c} =\frac{\sqrt c-4}{\sqrt c}\cdot\frac{2c}{c-16}$$

   $$=\frac{(\sqrt c-4)\cdot 2\sqrt c\cdot \sqrt c}{\sqrt c\cdot(\sqrt c-4)(\sqrt c+4)} =\frac{2\sqrt c}{\sqrt c+4}.$$

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

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

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

6. $$\left(\frac{\sqrt c+7}{\sqrt c-7}-\frac{\sqrt c-7}{\sqrt c+7}\right):\frac{14}{c-7\sqrt c}$$

   $$=\frac{(\sqrt c+7)^2-(\sqrt c-7)^2}{(\sqrt c-7)(\sqrt c+7)}\cdot\frac{c-7\sqrt c}{14}$$

   $$=\frac{28\sqrt c}{(\sqrt c-7)(\sqrt c+7)}\cdot\frac{\sqrt c(\sqrt c-7)}{14} =\frac{2c}{\sqrt c+7}.$$

7. $$\frac{\sqrt a-6}{a+3\sqrt a}-\frac{\sqrt a-3}{\sqrt a}+\frac{\sqrt a}{\sqrt a+3}$$

   $$=\frac{\sqrt a-6}{\sqrt a(\sqrt a+3)}-\frac{\sqrt a-3}{\sqrt a}+\frac{\sqrt a}{\sqrt a+3}$$

   $$=\frac{\sqrt a-6-(\sqrt a-3)(\sqrt a+3)+a}{\sqrt a(\sqrt a+3)} =\frac{\sqrt a+3}{\sqrt a(\sqrt a+3)} =\frac{1}{\sqrt a}.$$

8. $$\frac{\sqrt m+4}{m-6\sqrt m+9}:\frac{m-16}{2\sqrt m-6}-\frac{2}{\sqrt m-4}$$

   $$=\frac{\sqrt m+4}{(\sqrt m-3)^2}:\frac{(\sqrt m-4)(\sqrt m+4)}{2(\sqrt m-3)}-\frac{2}{\sqrt m-4}$$

   $$=\frac{2(\sqrt m-3)}{(\sqrt m-3)^2(\sqrt m-4)}-\frac{2}{\sqrt m-4} =\frac{2-2(\sqrt m-3)}{(\sqrt m-3)(\sqrt m-4)}-\frac{2}{\sqrt m-4}$$

   $$=\frac{8-2\sqrt m}{(\sqrt m-3)(\sqrt m-4)} =\frac{2(4-\sqrt m)}{(\sqrt m-3)(\sqrt m-4)} =-\frac{2}{\sqrt m-3} =\frac{2}{3-\sqrt m}.$$

Ответ

1) $$\frac{18}{a-3\sqrt a}$$; 2) $$\frac{5\sqrt b}{b-25}$$; 3) $$\frac{2(\sqrt m-\sqrt n)}{m}$$; 4) $$\frac{2\sqrt c}{\sqrt c+4}$$; 5) $$-\frac{a}{b}$$; 6) $$\frac{2c}{\sqrt c+7}$$; 7) $$\frac{1}{\sqrt a}$$; 8) $$\frac{2}{3-\sqrt m}$$.