вправа 5.23 гдз 10 клас математика Мерзляк Номіровський 2018
Вправа 5.23
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Відповідь ГДЗ:
\begin{equation} 1)\frac{\sqrt{a}-\sqrt{b}}{\sqrt[4]{a}+\sqrt[4]{b}}= \end{equation} \begin{equation} =\frac{ \left ( \sqrt[4]{a} \right )^{2}- \left ( \sqrt[4]{b} \right )^{2}}{\sqrt[4]{a}+\sqrt[4]{b}}= \end{equation} \begin{equation} =\frac{ \left ( \sqrt[4]{a}-\sqrt[4]{b} \right ) \left ( \sqrt[4]{a}+\sqrt[4]{b} \right )}{\sqrt[4]{a}+\sqrt[4]{b}}= \end{equation} \begin{equation} =\sqrt[4]{a}-\sqrt[4]{b}, \end{equation} \begin{equation} a > 0, b > 0; \end{equation} \begin{equation} 2)\frac{\sqrt[6]{x}-9}{\sqrt[12]{x}+3}= \end{equation} \begin{equation} =\frac{\left ( \sqrt[12]{x}-3 \right ) \left ( \sqrt[12]{x}+3 \right )}{\sqrt[12]{x}+3}= \end{equation} \begin{equation} =\sqrt[12]{x}-3, \end{equation} \begin{equation} x \geq 0; \end{equation} \begin{equation} 3)\frac{\sqrt[8]{ab^{2}}-\sqrt[8]{a^{2}b}}{\sqrt[4]{a}-\sqrt[4]{b}}= \end{equation} \begin{equation} =\frac{\sqrt[8]{ab} \left ( \sqrt[8]{b^{2}}-\sqrt[8]{a^{2}} \right )}{\sqrt[4]{a}-\sqrt[4]{b}}= \end{equation} \begin{equation} =\frac{\sqrt[8]{ab} \left ( \sqrt[4]{b}-\sqrt[4]{a} \right )}{\sqrt[4]{a}-\sqrt[4]{b}}= \end{equation} \begin{equation} =-\sqrt[8]{ab^{2}}, \end{equation} \begin{equation} a > 0, b > 0; a \neq b; \end{equation} \begin{equation} 4)\frac{\sqrt[3]{x^{2}}+4\sqrt[8]{x}+16}{x-64}= \end{equation} \begin{equation} =\frac{\sqrt[3]{x^{2}}+4\sqrt[8]{x}+16}{\left ( \sqrt[3]{x}-4 \right ) \left ( \sqrt[3]{x^{2}}+4\sqrt[3]{x}+16 \right )} = \end{equation} \begin{equation} =\frac{1}{\sqrt[3]{x}-4}, \end{equation} \begin{equation} x \neq 64; \end{equation} \begin{equation} 5)\frac{\sqrt{a}+\sqrt{b}}{\sqrt[3]{a}-\sqrt[6]{ab}+\sqrt[3]{b}}= \end{equation} \begin{equation} = \frac{\left ( \sqrt[6]{a}+\sqrt[6]{b} \right ) \left ( \sqrt[3]{a}-\sqrt[6]{ab}+\sqrt[3]{b} \right )}{\sqrt[3]{a}-\sqrt[6]{ab}+\sqrt[3]{b}}= \end{equation} \begin{equation} =\sqrt[6]{a}+\sqrt[6]{b}, \end{equation} \begin{equation} a > 0, b > 0; \end{equation} \begin{equation} 6) \frac{\sqrt[4]{a^{3}}-\sqrt[4]{a}+\sqrt{a}-1}{a-\sqrt{a}}= \end{equation} \begin{equation} =\frac{\sqrt[4]{a} \left ( \sqrt{a}-1 \right ) + \left ( \sqrt{a}-1 \right )}{\sqrt{a} \left ( \sqrt{a}-1 \right )}= \end{equation} \begin{equation} =\frac{ \left ( \sqrt{a}-1 \right ) \left ( \sqrt[4]{a}+1 \right )}{\sqrt{a} \left ( \sqrt{a}-1 \right )}= \end{equation} \begin{equation} =\frac{\sqrt[4]{a}+1}{\sqrt{a}}, \end{equation} \begin{equation} a > 0, a \neq 1. \end{equation}
\begin{equation} 1)\frac{\sqrt{a}-\sqrt{b}}{\sqrt[4]{a}+\sqrt[4]{b}}= \end{equation} \begin{equation} =\frac{ \left ( \sqrt[4]{a} \right )^{2}- \left ( \sqrt[4]{b} \right )^{2}}{\sqrt[4]{a}+\sqrt[4]{b}}= \end{equation} \begin{equation} =\frac{ \left ( \sqrt[4]{a}-\sqrt[4]{b} \right ) \left ( \sqrt[4]{a}+\sqrt[4]{b} \right )}{\sqrt[4]{a}+\sqrt[4]{b}}= \end{equation} \begin{equation} =\sqrt[4]{a}-\sqrt[4]{b}, \end{equation} \begin{equation} a > 0, b > 0; \end{equation} \begin{equation} 2)\frac{\sqrt[6]{x}-9}{\sqrt[12]{x}+3}= \end{equation} \begin{equation} =\frac{\left ( \sqrt[12]{x}-3 \right ) \left ( \sqrt[12]{x}+3 \right )}{\sqrt[12]{x}+3}= \end{equation} \begin{equation} =\sqrt[12]{x}-3, \end{equation} \begin{equation} x \geq 0; \end{equation} \begin{equation} 3)\frac{\sqrt[8]{ab^{2}}-\sqrt[8]{a^{2}b}}{\sqrt[4]{a}-\sqrt[4]{b}}= \end{equation} \begin{equation} =\frac{\sqrt[8]{ab} \left ( \sqrt[8]{b^{2}}-\sqrt[8]{a^{2}} \right )}{\sqrt[4]{a}-\sqrt[4]{b}}= \end{equation} \begin{equation} =\frac{\sqrt[8]{ab} \left ( \sqrt[4]{b}-\sqrt[4]{a} \right )}{\sqrt[4]{a}-\sqrt[4]{b}}= \end{equation} \begin{equation} =-\sqrt[8]{ab^{2}}, \end{equation} \begin{equation} a > 0, b > 0; a \neq b; \end{equation} \begin{equation} 4)\frac{\sqrt[3]{x^{2}}+4\sqrt[8]{x}+16}{x-64}= \end{equation} \begin{equation} =\frac{\sqrt[3]{x^{2}}+4\sqrt[8]{x}+16}{\left ( \sqrt[3]{x}-4 \right ) \left ( \sqrt[3]{x^{2}}+4\sqrt[3]{x}+16 \right )} = \end{equation} \begin{equation} =\frac{1}{\sqrt[3]{x}-4}, \end{equation} \begin{equation} x \neq 64; \end{equation} \begin{equation} 5)\frac{\sqrt{a}+\sqrt{b}}{\sqrt[3]{a}-\sqrt[6]{ab}+\sqrt[3]{b}}= \end{equation} \begin{equation} = \frac{\left ( \sqrt[6]{a}+\sqrt[6]{b} \right ) \left ( \sqrt[3]{a}-\sqrt[6]{ab}+\sqrt[3]{b} \right )}{\sqrt[3]{a}-\sqrt[6]{ab}+\sqrt[3]{b}}= \end{equation} \begin{equation} =\sqrt[6]{a}+\sqrt[6]{b}, \end{equation} \begin{equation} a > 0, b > 0; \end{equation} \begin{equation} 6) \frac{\sqrt[4]{a^{3}}-\sqrt[4]{a}+\sqrt{a}-1}{a-\sqrt{a}}= \end{equation} \begin{equation} =\frac{\sqrt[4]{a} \left ( \sqrt{a}-1 \right ) + \left ( \sqrt{a}-1 \right )}{\sqrt{a} \left ( \sqrt{a}-1 \right )}= \end{equation} \begin{equation} =\frac{ \left ( \sqrt{a}-1 \right ) \left ( \sqrt[4]{a}+1 \right )}{\sqrt{a} \left ( \sqrt{a}-1 \right )}= \end{equation} \begin{equation} =\frac{\sqrt[4]{a}+1}{\sqrt{a}}, \end{equation} \begin{equation} a > 0, a \neq 1. \end{equation}