ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 849

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\[1)\ f(x) = \frac{1 + \cos x}{\sin x}\]

\[= \frac{- \sin^{2}x - \cos x - \cos^{2}x}{\sin^{2}x} =\]

\[= - \frac{\left( \sin^{2}x + \cos^{2}x \right) + \cos x}{\sin^{2}x} =\]

\[= - \frac{1 + \cos x}{\sin^{2}x}.\]

\[2)\ f(x) = \frac{\sqrt{3x}}{3^{x} + 1}\]

\[= \frac{\frac{3 \bullet \left( 3^{x} + 1 \right)}{2\sqrt{3x}} - \sqrt{3x} \bullet 3^{x} \bullet \ln 3}{\left( 3^{x} + 1 \right)^{2}} =\]

\[= \frac{3 \bullet \left( 3^{x} + 1 \right) - 2 \bullet 3x \bullet 3^{x} \bullet \ln 3}{2\sqrt{3x} \bullet \left( 3^{x} + 1 \right)^{2}} =\]

\[= \frac{\sqrt{3} \bullet \left( 3^{x} + 1 \right) - 2x\sqrt{3} \bullet 3^{x} \bullet \ln 3}{2\sqrt{x} \bullet \left( 3^{x} + 1 \right)^{2}}.\]

\[3)\ f(x) = \frac{e^{0,5x}}{\cos{2x} - 5}\]

\[= \frac{0,5 \bullet e^{0,5x} \bullet \left( \cos{2x} - 5 + 4\sin{2x} \right)}{\left( \cos{2x} - 5 \right)^{2}}.\]

\[4)\ f(x) = \frac{5^{2x}}{\sin{3x} + 7}\]