Решебник по алгебре 9 класс Макарычев Задание 198

\[\boxed{\text{198}\text{\ (198)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ u = t^{\frac{2}{3}} + 1 \Longrightarrow t^{\frac{2}{3}} = u - 1;\]

\[v = t^{- \frac{2}{3}} + 1 \Longrightarrow t^{\frac{2}{3}} = v - 1 \Longrightarrow\]

\[\Longrightarrow \frac{1}{t^{\frac{2}{3}}} = v - 1 \Longrightarrow t^{\frac{2}{3}} = \frac{1}{v - 1};\]

\[u - 1 = \frac{1}{v - 1}\]

\[u = \frac{1}{v - 1} + 1.\]

\[\textbf{б)}\ u = (t + 2)^{\frac{1}{2}} \Longrightarrow u^{4} =\]

\[= t + 2 \Longrightarrow t = u^{4} - 2;\]

\[v = (2 - t)^{\frac{1}{4}} \Longrightarrow v^{4} =\]

\[= 2 - t \Longrightarrow t = 2 - v^{4}.\]

\[u^{4} - 2 = 2 - v^{4}\]

\[u^{4} = 4 - v^{4}.\]

\[\boxed{\mathbf{198}\text{.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ f(x) = |x^{2} - 2x|\]

\[\textbf{б)}\ f(x) = x² - 2|x|\]