Упр.16.70 ГДЗ Мордкович 8 класс (Алгебра)

1. а)
   $$\frac{a}{\sqrt{a}-3}-\frac{9}{\sqrt{a}-3}=\frac{a-9}{\sqrt{a}-3}$$
   $$\frac{a-9}{\sqrt{a}-3}=\frac{(\sqrt{a}-3)(\sqrt{a}+3)}{\sqrt{a}-3}=\sqrt{a}+3.$$

2. б)
   $$\frac{c}{\sqrt{c}-10}-\frac{20\sqrt{c}-100}{\sqrt{c}-10}=\frac{c-20\sqrt{c}+100}{\sqrt{c}-10}$$
   $$\frac{c-20\sqrt{c}+100}{\sqrt{c}-10}=\frac{(\sqrt{c}-10)^2}{\sqrt{c}-10}=\sqrt{c}-10.$$

3. в)
   $$\frac{c}{\sqrt{c}+9}-\frac{81}{\sqrt{c}+9}=\frac{c-81}{\sqrt{c}+9}$$
   $$\frac{c-81}{\sqrt{c}+9}=\frac{(\sqrt{c}-9)(\sqrt{c}+9)}{\sqrt{c}+9}=\sqrt{c}-9.$$

4. г)
   $$\frac{d}{\sqrt{d}+7}+\frac{14\sqrt{d}+49}{\sqrt{d}+7}=\frac{d+14\sqrt{d}+49}{\sqrt{d}+7}$$
   $$\frac{d+14\sqrt{d}+49}{\sqrt{d}+7}=\frac{(\sqrt{d}+7)^2}{\sqrt{d}+7}=\sqrt{d}+7.$$

Ответ

а) $$\sqrt{a}+3$$

б) $$\sqrt{c}-10$$

в) $$\sqrt{c}-9$$

г) $$\sqrt{d}+7$$