ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 21

\[1)\ y = \sin\sqrt{|x|}\]

\[\sin\sqrt{|x|} = 0\]

\[\sqrt{|x|} = \pi n\]

\[|x| = \pi^{2}n^{2}\]

\[x = \pm \pi^{2}n^{2}.\]

\[Найдем\ разность:\]

\[\pi^{2} \bullet 3^{2} = 9\pi^{2}\]

\[\pi^{2} \bullet 2^{2} = 4\pi^{2}\]

\[\pi^{2} \bullet 1^{2} = \pi^{2}\]

\[T_{1} = 9\pi^{2} - 4\pi^{2} = 5\pi^{2};\]

\[T_{2} = 4\pi^{2} - \pi^{2} = 3\pi^{2}.\]

\[Что\ и\ требовалось\ доказать.\]

\[2)\ y = \sin x + \sin{\sqrt{2}x}\]

\[y(x + T) = y(x);\]

\[\sin(x + T) + \sin\left( \sqrt{2}x + \sqrt{2}T \right) =\]

\[= \sin x + \sin{\sqrt{2}x}\]

\[T_{1} = 2\pi;\ \ \ T_{2} = \frac{2\pi}{\sqrt{2}};\ \ \ \]

\[nT_{1} = kT_{2}.\]

\[2\pi n = \frac{2\pi k}{\sqrt{2}}\]

\[n = \frac{k}{\sqrt{2}} \notin Z.\]

\[Что\ и\ требовалось\ доказать.\]