ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 213

\[1)\ y = \frac{1 + \cos x}{\sin x};\]

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

\[= \frac{- \left( 1 + \cos x \right)}{1 - \cos^{2}x} = \frac{- 1}{1 - \cos x} =\]

\[= \frac{1}{\cos x - 1}.\]

\[2)\ y = \frac{\sqrt{3x}}{3^{x} + 1};\]

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

\[3)\ y = \frac{x^{2} - 2x + 3}{x^{2} + 4x + 1};\]

\[= \frac{6x^{2} - 4x - 14}{\left( x^{2} + 4x + 1 \right)^{2}} =\]

\[= \frac{2\left( 3x^{2} - 2x - 7 \right)}{\left( x^{2} + 4x + 1 \right)^{2}}.\]

\[4)\ y = \frac{x^{2} - x + 1}{x^{2} + x + 1};\]

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