Упр.16.42 ГДЗ Мордкович 8 класс (Алгебра)

1. $$\frac{5}{\sqrt{x}+\sqrt{y}}=\frac{5(\sqrt{x}-\sqrt{y})}{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}=\frac{5(\sqrt{x}-\sqrt{y})}{x-y}.$$

2. $$\frac{1}{(\sqrt{a}-\sqrt{b})^2}=\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2(\sqrt{a}+\sqrt{b})^2}=\frac{(\sqrt{a}+\sqrt{b})^2}{\bigl((\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})\bigr)^2}=\frac{(\sqrt{a}+\sqrt{b})^2}{(a-b)^2}.$$

3. $$\frac{3}{\sqrt{m}-\sqrt{n}}=\frac{3(\sqrt{m}+\sqrt{n})}{(\sqrt{m}-\sqrt{n})(\sqrt{m}+\sqrt{n})}=\frac{3(\sqrt{m}+\sqrt{n})}{m-n}.$$

4. $$\frac{6}{(\sqrt{p}+\sqrt{q})^3}=\frac{6(\sqrt{p}-\sqrt{q})^3}{(\sqrt{p}+\sqrt{q})^3(\sqrt{p}-\sqrt{q})^3}=\frac{6(\sqrt{p}-\sqrt{q})^3}{\bigl((\sqrt{p}+\sqrt{q})(\sqrt{p}-\sqrt{q})\bigr)^3}=\frac{6(\sqrt{p}-\sqrt{q})^3}{(p-q)^3}.$$

Ответ

$$\frac{5(\sqrt{x}-\sqrt{y})}{x-y},\quad \frac{(\sqrt{a}+\sqrt{b})^2}{(a-b)^2},\quad \frac{3(\sqrt{m}+\sqrt{n})}{m-n},\quad \frac{6(\sqrt{p}-\sqrt{q})^3}{(p-q)^3}.$$