Упр.49 Повторение ГДЗ Колмогоров 10-11 класс (Алгебра)

а) (vk-(k^(3/4)+1)/(k^(1/4)+1))^(-1)-(k^(3/4)+vk)/(vk-1);
б) (((va+vb)^2-(2vb)^2)/(a-b)-(va-vb)/(va+vb)):32bvb/(va+vb);
в) ((x^(3/4)-y^(3/4))/(vx-vy)-(x^(1/4)+y^(1/4)))((x/y)^(1/4)+1);
г) (va^3+v(ab^2)-v(a^2b)-vb^3)/(b^(5/4)+(a^4b)^(1/4)-(ab^4)^(1/4)-a^(5/4)).

1. а)

   $$\left(\sqrt[4]{k}-\frac{\sqrt[4]{k^3}+1}{\sqrt[4]{k}+1}\right)^{-1}-\frac{\sqrt[4]{k^3}+\sqrt{k}}{\sqrt[4]{k}-1}$$
   $$=\left(\frac{\sqrt[4]{k^3}+\sqrt{k}-\sqrt[4]{k^3}-1}{\sqrt[4]{k}+1}\right)^{-1}-\frac{\sqrt{k}(\sqrt[4]{k}+1)}{(\sqrt[4]{k}+1)(\sqrt[4]{k}-1)}$$
   $$=\left(\frac{\sqrt[4]{k}-1}{\sqrt[4]{k}+1}\right)^{-1}-\frac{\sqrt{k}}{\sqrt[4]{k}-1}$$
   $$=\frac{\sqrt[4]{k}+1}{\sqrt[4]{k}-1}-\frac{\sqrt{k}}{\sqrt[4]{k}-1}$$
   $$=\frac{\sqrt[4]{k}+1-\sqrt{k}}{\sqrt[4]{k}-1}$$
   $$=\frac{1-\sqrt[4]{k}}{\sqrt[4]{k}-1}-1=-\sqrt[4]{k}-1.$$

2. б)

   $$\left(\frac{(\sqrt{a}+\sqrt{b})^2-(2\sqrt{b})^2}{a-b}-\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right):\frac{32b\sqrt{b}}{\sqrt{a}+\sqrt{b}}$$
   $$=\left(\frac{(\sqrt{a}-\sqrt{b})(\sqrt{a}+3\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}-\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right):\frac{32b\sqrt{b}}{\sqrt{a}+\sqrt{b}}$$
   $$=\frac{\sqrt{a}+3\sqrt{b}-(\sqrt{a}-\sqrt{b})}{\sqrt{a}+\sqrt{b}}:\frac{32b\sqrt{b}}{\sqrt{a}+\sqrt{b}}$$
   $$=\frac{4\sqrt{b}}{\sqrt{a}+\sqrt{b}}\cdot\frac{\sqrt{a}+\sqrt{b}}{32b\sqrt{b}}=\frac{1}{8b}.$$

3. в)

   $$\left(\frac{\sqrt[4]{x^3}-\sqrt[4]{y^3}}{\sqrt{x}-\sqrt{y}}-\left(\sqrt[4]{x}+\sqrt[4]{y}\right)\right)\left(\sqrt[4]{\frac{x}{y}}+1\right)$$
   $$=\left(\frac{(\sqrt[4]{x}-\sqrt[4]{y})(\sqrt{x}+\sqrt[4]{xy}+\sqrt{y})}{(\sqrt[4]{x}-\sqrt[4]{y})(\sqrt[4]{x}+\sqrt[4]{y})}-\left(\sqrt[4]{x}+\sqrt[4]{y}\right)\right)\left(\frac{\sqrt[4]{x}+\sqrt[4]{y}}{\sqrt[4]{y}}\right)$$
   $$=\frac{\sqrt{x}+\sqrt[4]{xy}+\sqrt{y}-\sqrt{x}-2\sqrt[4]{xy}-\sqrt{y}}{\sqrt[4]{x}+\sqrt[4]{y}}\cdot\frac{\sqrt[4]{x}+\sqrt[4]{y}}{\sqrt[4]{y}}$$
   $$=-\frac{\sqrt[4]{xy}}{\sqrt[4]{y}}=-\sqrt[4]{x}.$$

4. г)

   $$\frac{\sqrt{a^3}+\sqrt{ab^2}-\sqrt{a^2b}-\sqrt{b^3}}{\sqrt[4]{b^5}+\sqrt[4]{a^4b}-\sqrt[4]{ab^4}-\sqrt[4]{a^5}}$$
   $$=\frac{\sqrt{a}(a+b)-\sqrt{b}(a+b)}{\sqrt[4]{b}(b+a)-\sqrt[4]{a}(b+a)}$$
   $$=\frac{(\sqrt{a}-\sqrt{b})(a+b)}{(\sqrt[4]{b}-\sqrt[4]{a})(a+b)}$$
   $$=\frac{\sqrt{a}-\sqrt{b}}{\sqrt[4]{b}-\sqrt[4]{a}}.$$

   Обозначим $$u=\sqrt[4]{a},\ v=\sqrt[4]{b}.$$ Тогда
   $$\frac{\sqrt{a}-\sqrt{b}}{\sqrt[4]{b}-\sqrt[4]{a}}=\frac{u^2-v^2}{v-u}=\frac{(u-v)(u+v)}{v-u}=-(u+v).$$
   Значит,
   $$-\sqrt[4]{a}-\sqrt[4]{b}.$$

Ответ

а) $$-\sqrt[4]{k}-1$$; б) $$\frac{1}{8b}$$; в) $$-\sqrt[4]{x}$$; г) $$-\sqrt[4]{a}-\sqrt[4]{b}$$.