Решебник по алгебре 9 класс Макарычев Задание 572

 

\[\boxed{\text{572}\text{\ (572)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 4x^{4} + 4x² - 15 = 0\]

\[Пусть\ t = x^{2},\ t \geq 0,\ тогда:\]

\[4t^{2} + 4t - 15 = 0\]

\[D = 4 + 4 \cdot 15 = 64\]

\[t_{1,2} = \frac{- 2 \pm 8}{4},\]

\[так\ как\ t \geq 0,\ то\ t = 1,5.\]

\[\Longrightarrow x^{2} = 1,5 \Longrightarrow x = \pm \sqrt{1,5}.\]

\[Ответ:x = \pm \sqrt{1,5}.\]

\[\textbf{б)}\ 2x^{4} - x^{2} - 36 = 0\]

\[Пусть\ t = x^{2},\ t \geq 0,\ тогда:\]

\[2t^{2} - t - 36 = 0\]

\[D = 1 + 4 \cdot 2 \cdot 36 = 289\]

\[t_{1,2} = \frac{1 \pm 17}{4},\]

\[так\ как\ t \geq 0,\ то\ \ t = 4,5;\]

\[x² = 4,5 \Longrightarrow x = \pm \sqrt{4,5}.\]

\[Ответ:x = \pm \sqrt{4,5}.\]

\[\boxed{\text{572.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[S_{n} = \frac{\left( x_{1} + x_{n} \right) \cdot n}{2};\]

\[\textbf{а)}\ x_{n} = 4n + 2,\ \ \]

\[x_{1} = 4 + 2 = 6:\ \]

\[S_{n} = \frac{(4n + 2 + 6)}{2} \cdot n =\]

\[= 2n^{2} + 4n,\]

\[S_{50} = 2 \cdot 50^{2} + 4 \cdot 50 = 5200.\]

\[S_{100} = 2 \cdot 100^{2} + 4 \cdot 100 =\]

\[= 20\ 400.\]

\[\textbf{б)}\ x_{n} = 2n + 3,\ \ \]

\[x_{1} = 2 + 3 = 5:\ \]

\[S_{n} = \frac{(5 + 2n + 3)}{2} \cdot n = n^{2} + 4n;\]

\[S_{50} = 2500 + 200 = 2700;\]

\[S_{100} = 100² + 4 \cdot 100 = 10\ 400.\]

\[\textbf{в)}\ x_{n} = n - 4;\ \ \ x_{1} = 1 - 4 = - 3:\]

\[S_{n} = \frac{(n - 4 - 3)}{2} \cdot n = \frac{n^{2} - 7n}{2}\]

\[S_{50} = \frac{50^{2} - 7 \cdot 50}{2} =\]

\[= \frac{2500 - 350}{2} = \frac{2150}{2} = 1075.\]

\[S_{100} = \frac{100^{2} - 7 \cdot 100}{2} =\]

\[= \frac{10\ 000 - 700}{2} = \frac{9300}{2} = 4650.\]

\[\textbf{г)}\ \ x_{n} = 3n - 1;\ \ \ x_{1} = 3 - 1 =\]

\[= 2:\]

\[S_{n} = \frac{(3n - 1 + 2)}{2} \cdot n =\]

\[= \frac{3n^{2} + 1n}{2}\]

\[S_{50} = \frac{3 \cdot 50^{2} + 1 \cdot 50}{2} =\]

\[= \frac{3 \cdot 2500 + 50}{2} = \frac{7550}{2} = 3775.\]

\[S_{100} = \frac{3 \cdot 100^{2} + 1 \cdot 100}{2} =\]

\[= \frac{30100}{2} = 15050.\]