Упр.9.46 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1. Докажите тождество:
   1) $$\left(\frac{1}{x^{1/6}+1}-\frac{x^{1/6}-1}{x^{1/3}}\right):\frac{x^{2/3}}{x^{1/3}+2x^{1/6}+1}=\frac{x^{1/6}+1}{x};$$
   2) $$\frac{\frac{a+b}{a^{2/3}-b^{2/3}}+\frac{(ab^2)^{1/3}-(a^2b)^{1/3}}{a^{2/3}-2(ab)^{1/3}+b^{2/3}}}{a^{1/6}-b^{1/6}}=a^{1/6}+b^{1/6}.$$

1)

$$\left(\frac{1}{\sqrt[6]{x}+1}-\frac{\sqrt[6]{x}-1}{\sqrt[3]{x}}\right):\frac{\sqrt[3]{x}}{\sqrt[3]{x}+2\sqrt[6]{x}+1}$$
$$=\frac{\sqrt[3]{x}-(\sqrt[6]{x}-1)(\sqrt[6]{x}+1)}{\sqrt[3]{x}(\sqrt[6]{x}+1)}\cdot\frac{\sqrt[3]{x}+2\sqrt[6]{x}+1}{\sqrt[3]{x}}$$
$$=\frac{\sqrt[3]{x}-(\sqrt[3]{x}-1)(\sqrt[6]{x}+1)^2}{\sqrt[3]{x}(\sqrt[6]{x}+1)}\cdot\frac{\sqrt[3]{x}+2\sqrt[6]{x}+1}{\sqrt[3]{x}}$$
$$=\frac{1}{\sqrt[3]{x}}\cdot\frac{\sqrt[6]{x}+1}{\sqrt[3]{x}} =\frac{\sqrt[6]{x}+1}{x}.$$

Тождество доказано.

2)

$$\frac{\dfrac{a+b}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}+\dfrac{\sqrt[3]{ab^2}-\sqrt[3]{a^2b}}{\sqrt[3]{a^2}-2\sqrt[3]{ab}+\sqrt[3]{b^2}}}{\sqrt[6]{a}-\sqrt[6]{b}}$$
$$=\frac{\dfrac{\sqrt[3]{a^3}+\sqrt[3]{b^3}}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}+\dfrac{\sqrt[3]{ab}\,(\sqrt[3]{b}-\sqrt[3]{a})}{(\sqrt[3]{a}-\sqrt[3]{b})^2}}{\sqrt[6]{a}-\sqrt[6]{b}}$$
$$=\frac{(\sqrt[6]{a}+\sqrt[6]{b})(\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2})-\sqrt[3]{ab}}{\sqrt[3]{a}-\sqrt[3]{b}}$$
$$=\frac{\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}-\sqrt[3]{ab}}{\sqrt[3]{a}-\sqrt[3]{b}} =\frac{(\sqrt[3]{a}-\sqrt[3]{b})^2}{\sqrt[3]{a}-\sqrt[3]{b}} =\sqrt[3]{a}-\sqrt[3]{b}.$$

Тождество доказано.