Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 298

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\[= 5^{2} + \frac{10}{2} - 3^{3} =\]

\[= 25 + 5 - 27 = 3\]

\[= \left( \frac{3}{9^{\log_{9}4}} + \left( 125^{\log_{125}8} \right)^{\frac{2}{3}} \right) \bullet 2^{2} =\]

\[= \left( \frac{3}{4} + 8^{\frac{2}{3}} \right) \bullet 2^{2} =\]

\[= \left( \frac{3}{2^{2}} + \left( 2^{3} \right)^{\frac{2}{3}} \right) \bullet 2^{2} =\]

\[= \left( \frac{3}{2^{2}} + 2^{2} \right) \bullet 2^{2} = 3 + 2^{4} =\]

\[= 3 + 16 = 19\]

\[3)\ 16^{1 + \log_{4}5} + 4^{\frac{1}{2}\log_{2}3 + 3\log_{8}5} =\]

\[= 16 \bullet 5^{2} + 3 \bullet \left( 8^{\log_{8}5} \right)^{2} =\]

\[= 16 \bullet 25 + 3 \bullet 5^{2} =\]

\[= 400 + 3 \bullet 25 = 400 + 75 =\]

\[= 475\]

\[= 72 \bullet \left( \frac{7^{\log_{7}9}}{\left( 7^{\log_{7}6} \right)^{2}} + 4^{- 2} \right) =\]

\[= 72 \bullet \left( \frac{9}{6^{2}} + \frac{1}{4^{2}} \right) =\]

\[= 72 \bullet \left( \frac{9}{36} + \frac{1}{16} \right) =\]

\[= 72 \bullet \left( \frac{1}{4} + \frac{1}{16} \right) =\]

\[= 71 \bullet \left( \frac{4}{16} + \frac{1}{16} \right) = 72 \bullet \frac{5}{16} =\]

\[= 4,5 \bullet 5 = 22,5\]