Упр.22.19 ГДЗ Мерзляк Поляков 9 класс (Алгебра)
1) (va-vb)/(a^(1/4)+b^(1/4));
2) (x^(1/6)-9)/(x^(1/12)+3);
3) (vm+m^(1/4))/(m-(m^3)^(1/4));
4) ((ab^2)^(1/8)-(a^2 b)^(1/8))/(a^(1/4)-b^(1/4));
5) (a(b^2)^(1/3)-b(a^2)^(1/3))/(a^2 b^2)^(1/3);
6) ((x^2)^(1/3)+4x^(1/3)+16)/(x-64);
7) (va+vb)/(a^(1/3)-(ab)^(1/6)+b^(1/3));
8) (2-2^(1/3))/2^(1/3);
9) ((a^3)^(1/4)-a^(1/4)+va-1)/(a-va).
$$\frac{\sqrt a-\sqrt b}{\sqrt[4]{a}+\sqrt[4]{b}}= \frac{(\sqrt[4]{a})^2-(\sqrt[4]{b})^2}{\sqrt[4]{a}+\sqrt[4]{b}}= \frac{(\sqrt[4]{a}-\sqrt[4]{b})(\sqrt[4]{a}+\sqrt[4]{b})}{\sqrt[4]{a}+\sqrt[4]{b}}=\sqrt[4]{a}-\sqrt[4]{b}.$$
$$\frac{\sqrt[6]{x}-9}{\sqrt[12]{x}+3}= \frac{(\sqrt[12]{x})^2-3^2}{\sqrt[12]{x}+3}= \frac{(\sqrt[12]{x}-3)(\sqrt[12]{x}+3)}{\sqrt[12]{x}+3}=\sqrt[12]{x}-3.$$
$$\frac{\sqrt m+\sqrt[4]{m}}{m-\sqrt[4]{m^3}}= \frac{(\sqrt[4]{m})^2+\sqrt[4]{m}}{(\sqrt[4]{m})^4-(\sqrt[4]{m})^3}= \frac{\sqrt[4]{m}(\sqrt[4]{m}+1)}{(\sqrt[4]{m})^3(\sqrt[4]{m}-1)}= \frac{\sqrt[4]{m}+1}{\sqrt[4]{m^3}-\sqrt m}.$$
$$\frac{\sqrt[8]{ab^2}-\sqrt[8]{a^2b}}{\sqrt[4]{a}-\sqrt[4]{b}}= \frac{\sqrt[8]{ab}\,(\sqrt[8]{b}-\sqrt[8]{a})}{(\sqrt[8]{a})^2-(\sqrt[8]{b})^2}= \frac{-\sqrt[8]{ab}\,(\sqrt[8]{a}-\sqrt[8]{b})}{(\sqrt[8]{a}-\sqrt[8]{b})(\sqrt[8]{a}+\sqrt[8]{b})} =-\frac{\sqrt[8]{ab}}{\sqrt[8]{a}+\sqrt[8]{b}}.$$
$$\frac{a\sqrt[3]{b^2}-b\sqrt[3]{a^2}}{\sqrt[3]{a^2b^2}}= \frac{\sqrt[3]{a^3b^2}-\sqrt[3]{a^2b^3}}{\sqrt[3]{a^2b^2}}= \frac{\sqrt[3]{a^2b^2}(\sqrt[3]{a}-\sqrt[3]{b})}{\sqrt[3]{a^2b^2}}= \sqrt[3]{a}-\sqrt[3]{b}.$$
$$\frac{\sqrt[3]{x^2}+4\sqrt[3]{x}+16}{x-64}= \frac{(\sqrt[3]{x})^2+4\sqrt[3]{x}+4^2}{(\sqrt[3]{x})^3-4^3}= \frac{(\sqrt[3]{x}+4)(\sqrt[3]{x}-4)}{(\sqrt[3]{x}-4)\bigl((\sqrt[3]{x})^2+4\sqrt[3]{x}+16\bigr)}= \frac{1}{\sqrt[3]{x}-4}.$$
$$\frac{\sqrt a+\sqrt b}{\sqrt[3]{a}-\sqrt[6]{ab}+\sqrt[3]{b}}= \frac{(\sqrt a+\sqrt b)(\sqrt[6]{a}+\sqrt[6]{b})}{(\sqrt[6]{a})^3-(\sqrt[6]{b})^3}= \frac{(\sqrt a+\sqrt b)(\sqrt[6]{a}+\sqrt[6]{b})}{(\sqrt a+\sqrt b)(\sqrt[6]{a}+\sqrt[6]{b})} =\sqrt[6]{a}+\sqrt[6]{b}.$$
$$\frac{2-\sqrt[3]{2}}{\sqrt[3]{2}}= \frac{\sqrt[3]{2^3}-\sqrt[3]{2}}{\sqrt[3]{2}}= \frac{\sqrt[3]{2}\,(\sqrt[3]{4}-1)}{\sqrt[3]{2}}=\sqrt[3]{4}-1.$$
$$\frac{\sqrt[4]{a^3}-\sqrt[4]{a}+\sqrt a-1}{a-\sqrt a}= \frac{\sqrt[4]{a}\,(\sqrt[4]{a^2}-1)+(\sqrt a-1)}{(\sqrt a)^2-\sqrt a}= \frac{(\sqrt[4]{a}+1)(\sqrt a-1)}{\sqrt a(\sqrt a-1)}= \frac{\sqrt[4]{a}+1}{\sqrt a}.$$
Ответ
- $$\sqrt[4]{a}-\sqrt[4]{b}$$
- $$\sqrt[12]{x}-3$$
- $$\frac{\sqrt[4]{m}+1}{\sqrt[4]{m^3}-\sqrt m}$$
- $$-\frac{\sqrt[8]{ab}}{\sqrt[8]{a}+\sqrt[8]{b}}$$
- $$\sqrt[3]{a}-\sqrt[3]{b}$$
- $$\frac{1}{\sqrt[3]{x}-4}$$
- $$\sqrt[6]{a}+\sqrt[6]{b}$$
- $$\sqrt[3]{4}-1$$
- $$\frac{\sqrt[4]{a}+1}{\sqrt a}$$