вправа 26.15 гдз 10 клас математика Мерзляк Номіровський 2018
Вправа 26.15
Умова:
Умова:
Доведіть тотожність.
Відповідь ГДЗ:
1) Спрощуємо ліву частину тотожності: \begin{equation} \left (\frac{m-n}{m^{\frac{3}{4}}+m^{\frac{1}{2}}n^{\frac{1}{4}}}-\frac{m^{\frac{1}{2}}-n^{\frac{1}{2}}}{m^{\frac{1}{4}}+n^{\frac{1}{4}}} \right ) \cdot \end{equation} \begin{equation} \cdot \left (\frac{n}{m} \right )^{-\frac{1}{2}}= \end{equation} \begin{equation} =\frac{\left (m-n-m+m^{\frac{1}{2}}n^{\frac{1}{2}} \right )m^{\frac{1}{2}}}{m^{\frac{1}{2}}\left (m^{\frac{1}{4}}+n^{\frac{1}{4}} \right )n^{\frac{1}{2}}}= \end{equation} \begin{equation} =\frac{n^{\frac{1}{2}}\left (m^{\frac{1}{2}}-n^{\frac{1}{2}} \right )}{\left (m^{\frac{1}{4}}+n^{\frac{1}{4}} \right )n^{\frac{1}{2}}}= \end{equation} \begin{equation} =\frac{\left (m^{\frac{1}{4}}-n^{\frac{1}{4}} \right ) \left (m^{\frac{1}{4}}+n^{\frac{1}{4}} \right )}{\left (m^{\frac{1}{4}}+n^{\frac{1}{4}} \right )}= \end{equation} \begin{equation} =m^{\frac{1}{4}}-n^{\frac{1}{4}}; \end{equation} \begin{equation} m^{\frac{1}{4}}-n^{\frac{1}{4}}=m^{\frac{1}{4}}-n^{\frac{1}{4}}. \end{equation} Доведено.
2) Спрощуємо ліву частину тотожності: \begin{equation} \frac{a+b}{a^{\frac{2}{3}}-a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}}}- \end{equation} \begin{equation} -\frac{a-b}{a^{\frac{2}{3}}+a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}}}-\frac{a^{\frac{2}{3}}-b^{\frac{2}{3}}}{a^{\frac{1}{3}}-b^{\frac{1}{3}}}= \end{equation} \begin{equation} =\frac{\left (a^{\frac{1}{3}}-b^{\frac{1}{3}} \right ) \left (a^{\frac{2}{3}}-a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}} \right )}{\left (a^{\frac{2}{3}}-a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}} \right )}- \end{equation} \begin{equation} -\frac{\left (a^{\frac{1}{3}}-b^{\frac{1}{3}} \right ) \left (a^{\frac{2}{3}}+a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}} \right )}{\left (a^{\frac{2}{3}}+a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}} \right )}- \end{equation} \begin{equation} \frac{\left (a^{\frac{1}{3}}-b^{\frac{1}{3}} \right ) \left (a^{\frac{1}{3}}+b^{\frac{1}{3}} \right )}{\left (a^{\frac{1}{3}}-b^{\frac{1}{3}} \right )}= \end{equation} \begin{equation} =a^{\frac{1}{3}}+b^{\frac{1}{3}}-a^{\frac{1}{3}}+ \end{equation} \begin{equation} +b^{\frac{1}{3}}-a^{\frac{1}{3}}-b^{\frac{1}{3}}= \end{equation} \begin{equation} =b^{\frac{1}{3}}-a^{\frac{1}{3}}; \end{equation} \begin{equation} b^{\frac{1}{3}}-a^{\frac{1}{3}}= \end{equation} \begin{equation} =b^{\frac{1}{3}}-a^{\frac{1}{3}}. \end{equation} Доведено.
1) Спрощуємо ліву частину тотожності: \begin{equation} \left (\frac{m-n}{m^{\frac{3}{4}}+m^{\frac{1}{2}}n^{\frac{1}{4}}}-\frac{m^{\frac{1}{2}}-n^{\frac{1}{2}}}{m^{\frac{1}{4}}+n^{\frac{1}{4}}} \right ) \cdot \end{equation} \begin{equation} \cdot \left (\frac{n}{m} \right )^{-\frac{1}{2}}= \end{equation} \begin{equation} =\frac{\left (m-n-m+m^{\frac{1}{2}}n^{\frac{1}{2}} \right )m^{\frac{1}{2}}}{m^{\frac{1}{2}}\left (m^{\frac{1}{4}}+n^{\frac{1}{4}} \right )n^{\frac{1}{2}}}= \end{equation} \begin{equation} =\frac{n^{\frac{1}{2}}\left (m^{\frac{1}{2}}-n^{\frac{1}{2}} \right )}{\left (m^{\frac{1}{4}}+n^{\frac{1}{4}} \right )n^{\frac{1}{2}}}= \end{equation} \begin{equation} =\frac{\left (m^{\frac{1}{4}}-n^{\frac{1}{4}} \right ) \left (m^{\frac{1}{4}}+n^{\frac{1}{4}} \right )}{\left (m^{\frac{1}{4}}+n^{\frac{1}{4}} \right )}= \end{equation} \begin{equation} =m^{\frac{1}{4}}-n^{\frac{1}{4}}; \end{equation} \begin{equation} m^{\frac{1}{4}}-n^{\frac{1}{4}}=m^{\frac{1}{4}}-n^{\frac{1}{4}}. \end{equation} Доведено.
2) Спрощуємо ліву частину тотожності: \begin{equation} \frac{a+b}{a^{\frac{2}{3}}-a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}}}- \end{equation} \begin{equation} -\frac{a-b}{a^{\frac{2}{3}}+a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}}}-\frac{a^{\frac{2}{3}}-b^{\frac{2}{3}}}{a^{\frac{1}{3}}-b^{\frac{1}{3}}}= \end{equation} \begin{equation} =\frac{\left (a^{\frac{1}{3}}-b^{\frac{1}{3}} \right ) \left (a^{\frac{2}{3}}-a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}} \right )}{\left (a^{\frac{2}{3}}-a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}} \right )}- \end{equation} \begin{equation} -\frac{\left (a^{\frac{1}{3}}-b^{\frac{1}{3}} \right ) \left (a^{\frac{2}{3}}+a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}} \right )}{\left (a^{\frac{2}{3}}+a^{\frac{1}{3}}b^{\frac{1}{3}}+b^{\frac{2}{3}} \right )}- \end{equation} \begin{equation} \frac{\left (a^{\frac{1}{3}}-b^{\frac{1}{3}} \right ) \left (a^{\frac{1}{3}}+b^{\frac{1}{3}} \right )}{\left (a^{\frac{1}{3}}-b^{\frac{1}{3}} \right )}= \end{equation} \begin{equation} =a^{\frac{1}{3}}+b^{\frac{1}{3}}-a^{\frac{1}{3}}+ \end{equation} \begin{equation} +b^{\frac{1}{3}}-a^{\frac{1}{3}}-b^{\frac{1}{3}}= \end{equation} \begin{equation} =b^{\frac{1}{3}}-a^{\frac{1}{3}}; \end{equation} \begin{equation} b^{\frac{1}{3}}-a^{\frac{1}{3}}= \end{equation} \begin{equation} =b^{\frac{1}{3}}-a^{\frac{1}{3}}. \end{equation} Доведено.