Упр.947 ГДЗ Никольский Потапов 9 класс (Алгебра)

а) (v15-v6)/(v35-v14);
б) (v10+v15)/(v8+v12);
в) (((x^2)y)^(1/3)-(xy^2)^(1/3))/((ax)^(1/3)-(ay)^(1/3));
г) (n(m)^(1/3)-m(n)^(1/3))/((m^2)^(1/3)+(n^2)^(1/3)+2(mn)^(1/3)).

1. а)

   $$\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}= \frac{\sqrt{3}\,(\sqrt{5}-\sqrt{2})}{\sqrt{7}\,(\sqrt{5}-\sqrt{2})}= \frac{\sqrt{3}}{\sqrt{7}}= \frac{\sqrt{21}}{7}.$$

2. б)

   $$\frac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}= \frac{\sqrt{5}\,(\sqrt{2}+\sqrt{3})}{\sqrt{4}\,(\sqrt{2}+\sqrt{3})}= \frac{\sqrt{5}}{2}.$$

3. в)

   $$\frac{\sqrt[3]{x^2y}-\sqrt[3]{xy^2}}{\sqrt[3]{ax}-\sqrt[3]{ay}}= \frac{\sqrt[3]{xy}\,(\sqrt[3]{x}-\sqrt[3]{y})}{\sqrt[3]{a}\,(\sqrt[3]{x}-\sqrt[3]{y})}= \frac{\sqrt[3]{xy}}{\sqrt[3]{a}}= \sqrt[3]{\frac{xy}{a}}.$$

4. г)

   $$\frac{n\sqrt[3]{m}-m\sqrt[3]{n}}{\sqrt[3]{m^2}+\sqrt[3]{n^2}+2\sqrt[3]{mn}}= \frac{\sqrt[3]{mn}\,(\sqrt[3]{n}-\sqrt[3]{m})}{(\sqrt[3]{m}+\sqrt[3]{n})^2}.$$

   Так как
   $$\sqrt[3]{m^2}+\sqrt[3]{n^2}+2\sqrt[3]{mn}=(\sqrt[3]{m}+\sqrt[3]{n})^2,$$
   то
   $$\frac{n\sqrt[3]{m}-m\sqrt[3]{n}}{\sqrt[3]{m^2}+\sqrt[3]{n^2}+2\sqrt[3]{mn}}= \frac{\sqrt[3]{mn}\,(\sqrt[3]{n}-\sqrt[3]{m})}{\sqrt[3]{m}+\sqrt[3]{n}}.$$

Ответ

а) $$\frac{\sqrt{21}}{7}$$; б) $$\frac{\sqrt{5}}{2}$$; в) $$\sqrt[3]{\frac{xy}{a}}$$; г) $$\frac{\sqrt[3]{mn}\,(\sqrt[3]{n}-\sqrt[3]{m})}{\sqrt[3]{m}+\sqrt[3]{n}}$$.