Упр.22.19 ГДЗ Мерзляк Поляков 9 класс (Алгебра)
Сократите дробь:
- $$\frac{\sqrt{a}-\sqrt{b}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}$$
- $$\frac{x^{\frac{1}{6}}-9}{x^{\frac{1}{12}}+3}$$
- $$\frac{\sqrt{m}+m^{\frac{1}{4}}}{m-\left(m^3\right)^{\frac{1}{4}}}$$
- $$\frac{\left(ab^2\right)^{\frac{1}{8}}-\left(a^2b\right)^{\frac{1}{8}}}{a^{\frac{1}{4}}-b^{\frac{1}{4}}}$$
- $$\frac{a\left(b^2\right)^{\frac{1}{3}}-b\left(a^2\right)^{\frac{1}{3}}}{\left(a^2b^2\right)^{\frac{1}{3}}}$$
- $$\frac{\left(x^2\right)^{\frac{1}{3}}+4x^{\frac{1}{3}}+16}{x-64}$$
- $$\frac{\sqrt{a}+\sqrt{b}}{a^{\frac{1}{3}}-(ab)^{\frac{1}{6}}+b^{\frac{1}{3}}}$$
- $$\frac{2-\sqrt[3]{2}}{\sqrt[3]{2}}$$
- $$\frac{\left(a^3\right)^{\frac{1}{4}}-a^{\frac{1}{4}}+\sqrt{a}-1}{a-\sqrt{a}}$$
$$\frac{\sqrt a-\sqrt b}{\sqrt[4]{a}+\sqrt[4]{b}}= \frac{(\sqrt[4]{a})^2-(\sqrt[4]{b})^2}{\sqrt[4]{a}+\sqrt[4]{b}}= \frac{(\sqrt[4]{a}-\sqrt[4]{b})(\sqrt[4]{a}+\sqrt[4]{b})}{\sqrt[4]{a}+\sqrt[4]{b}}=\sqrt[4]{a}-\sqrt[4]{b}.$$
$$\frac{\sqrt[6]{x}-9}{\sqrt[12]{x}+3}= \frac{(\sqrt[12]{x})^2-3^2}{\sqrt[12]{x}+3}= \frac{(\sqrt[12]{x}-3)(\sqrt[12]{x}+3)}{\sqrt[12]{x}+3}=\sqrt[12]{x}-3.$$
$$\frac{\sqrt m+\sqrt[4]{m}}{m-\sqrt[4]{m^3}}= \frac{(\sqrt[4]{m})^2+\sqrt[4]{m}}{(\sqrt[4]{m})^4-(\sqrt[4]{m})^3}= \frac{\sqrt[4]{m}(\sqrt[4]{m}+1)}{(\sqrt[4]{m})^3(\sqrt[4]{m}-1)}= \frac{\sqrt[4]{m}+1}{\sqrt[4]{m^3}-\sqrt m}.$$
$$\frac{\sqrt[8]{ab^2}-\sqrt[8]{a^2b}}{\sqrt[4]{a}-\sqrt[4]{b}}= \frac{\sqrt[8]{ab}\,(\sqrt[8]{b}-\sqrt[8]{a})}{(\sqrt[8]{a})^2-(\sqrt[8]{b})^2}= \frac{-\sqrt[8]{ab}\,(\sqrt[8]{a}-\sqrt[8]{b})}{(\sqrt[8]{a}-\sqrt[8]{b})(\sqrt[8]{a}+\sqrt[8]{b})} =-\frac{\sqrt[8]{ab}}{\sqrt[8]{a}+\sqrt[8]{b}}.$$
$$\frac{a\sqrt[3]{b^2}-b\sqrt[3]{a^2}}{\sqrt[3]{a^2b^2}}= \frac{\sqrt[3]{a^3b^2}-\sqrt[3]{a^2b^3}}{\sqrt[3]{a^2b^2}}= \frac{\sqrt[3]{a^2b^2}(\sqrt[3]{a}-\sqrt[3]{b})}{\sqrt[3]{a^2b^2}}= \sqrt[3]{a}-\sqrt[3]{b}.$$
$$\frac{\sqrt[3]{x^2}+4\sqrt[3]{x}+16}{x-64}= \frac{(\sqrt[3]{x})^2+4\sqrt[3]{x}+4^2}{(\sqrt[3]{x})^3-4^3}= \frac{(\sqrt[3]{x}+4)(\sqrt[3]{x}-4)}{(\sqrt[3]{x}-4)\bigl((\sqrt[3]{x})^2+4\sqrt[3]{x}+16\bigr)}= \frac{1}{\sqrt[3]{x}-4}.$$
$$\frac{\sqrt a+\sqrt b}{\sqrt[3]{a}-\sqrt[6]{ab}+\sqrt[3]{b}}= \frac{(\sqrt a+\sqrt b)(\sqrt[6]{a}+\sqrt[6]{b})}{(\sqrt[6]{a})^3-(\sqrt[6]{b})^3}= \frac{(\sqrt a+\sqrt b)(\sqrt[6]{a}+\sqrt[6]{b})}{(\sqrt a+\sqrt b)(\sqrt[6]{a}+\sqrt[6]{b})} =\sqrt[6]{a}+\sqrt[6]{b}.$$
$$\frac{2-\sqrt[3]{2}}{\sqrt[3]{2}}= \frac{\sqrt[3]{2^3}-\sqrt[3]{2}}{\sqrt[3]{2}}= \frac{\sqrt[3]{2}\,(\sqrt[3]{4}-1)}{\sqrt[3]{2}}=\sqrt[3]{4}-1.$$
$$\frac{\sqrt[4]{a^3}-\sqrt[4]{a}+\sqrt a-1}{a-\sqrt a}= \frac{\sqrt[4]{a}\,(\sqrt[4]{a^2}-1)+(\sqrt a-1)}{(\sqrt a)^2-\sqrt a}= \frac{(\sqrt[4]{a}+1)(\sqrt a-1)}{\sqrt a(\sqrt a-1)}= \frac{\sqrt[4]{a}+1}{\sqrt a}.$$
Ответ
- $$\sqrt[4]{a}-\sqrt[4]{b}$$
- $$\sqrt[12]{x}-3$$
- $$\frac{\sqrt[4]{m}+1}{\sqrt[4]{m^3}-\sqrt m}$$
- $$-\frac{\sqrt[8]{ab}}{\sqrt[8]{a}+\sqrt[8]{b}}$$
- $$\sqrt[3]{a}-\sqrt[3]{b}$$
- $$\frac{1}{\sqrt[3]{x}-4}$$
- $$\sqrt[6]{a}+\sqrt[6]{b}$$
- $$\sqrt[3]{4}-1$$
- $$\frac{\sqrt[4]{a}+1}{\sqrt a}$$
