Упр.16.78 ГДЗ Мордкович 8 класс (Алгебра)

1. $$\frac{x-10\sqrt{x}+25}{3\sqrt{x}+12}:\frac{2\sqrt{x}-10}{x-16}$$
   $$=\frac{(\sqrt{x}-5)^2}{3(\sqrt{x}+4)}:\frac{2(\sqrt{x}-5)}{(\sqrt{x}-4)(\sqrt{x}+4)}$$
   $$=\frac{(\sqrt{x}-5)^2}{3(\sqrt{x}+4)}\cdot\frac{(\sqrt{x}-4)(\sqrt{x}+4)}{2(\sqrt{x}-5)}$$
   $$=\frac{(\sqrt{x}-5)(\sqrt{x}-4)}{6}.$$

2. $$\frac{1-a}{4\sqrt{a}+8\sqrt{b}}\cdot\frac{a+4\sqrt{ab}+4b}{3-3\sqrt{a}}$$
   $$=\frac{(1-\sqrt{a})(1+\sqrt{a})\cdot(\sqrt{a}+2\sqrt{b})^2}{4(\sqrt{a}+2\sqrt{b})\cdot 3(1-\sqrt{a})}$$
   $$=\frac{(1+\sqrt{a})(\sqrt{a}+2\sqrt{b})}{12}.$$

3. $$\frac{c-25}{c+12\sqrt{c}+36}\cdot\frac{3\sqrt{c}+18}{2\sqrt{c}+10}$$
   $$=\frac{(\sqrt{c}-5)(\sqrt{c}+5)\cdot 3(\sqrt{c}+6)}{(\sqrt{c}+6)^2\cdot 2(\sqrt{c}+5)}$$
   $$=\frac{3(\sqrt{c}-5)}{2(\sqrt{c}+6)}.$$

4. $$\frac{5\sqrt{m}-10\sqrt{n}}{\sqrt{m}-5}:\frac{4n-4\sqrt{mn}+m}{15-3\sqrt{m}}$$
   $$=\frac{5(\sqrt{m}-2\sqrt{n})}{\sqrt{m}-5}:\frac{(\sqrt{m}-2\sqrt{n})^2}{-3(\sqrt{m}-5)}$$
   $$=\frac{5(\sqrt{m}-2\sqrt{n})}{\sqrt{m}-5}\cdot\frac{-3(\sqrt{m}-5)}{(\sqrt{m}-2\sqrt{n})^2}$$
   $$=-\frac{15}{\sqrt{m}-2\sqrt{n}}=\frac{15}{2\sqrt{n}-\sqrt{m}}.$$

Ответ

а) $$\frac{(\sqrt{x}-5)(\sqrt{x}-4)}{6}$$;
б) $$\frac{(1+\sqrt{a})(\sqrt{a}+2\sqrt{b})}{12}$$;
в) $$\frac{3(\sqrt{c}-5)}{2(\sqrt{c}+6)}$$;
г) $$\frac{15}{2\sqrt{n}-\sqrt{m}}$$.