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МатематикаГДЗ по алгебре 9 класс Макарычев Задание 748
Задание 748
\[\boxed{\text{748\ (748).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\frac{15!}{14!} = \frac{14! \cdot 15!}{14!} = 15;\]
\[\textbf{б)}\frac{8!}{10!} = \frac{8!}{8! \cdot 9 \cdot 10} = \frac{1}{90};\]
\[\textbf{в)}\frac{42!}{40!} = \frac{40! \cdot 41 \cdot 42}{40!} = 41 \cdot 42 =\]
\[= 1722;\]
\[\textbf{г)}\ \frac{16!}{14! \cdot 3!} = \frac{14! \cdot 15 \cdot 16}{14! \cdot 3!} =\]
\[= \frac{3 \cdot 5 \cdot 2 \cdot 8}{1 \cdot 2 \cdot 3} = 40;\]
\[\textbf{д)}\ \frac{28!}{4! \cdot 26!} = \frac{26! \cdot 27 \cdot 28}{4! \cdot 26!} =\]
\[= \frac{3 \cdot 9 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3 \cdot 4} = \frac{63}{2} = 31,5;\]
\[\textbf{е)}\ \frac{45!}{43! \cdot 3!} = \frac{43! \cdot 44 \cdot 45}{43! \cdot 3!} =\]
\[= \frac{2 \cdot 22 \cdot 3 \cdot 15}{1 \cdot 2 \cdot 3} = 330.\]
\[\boxed{\text{748.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ 4x^{4} - 17x^{2} + 4 = 0\]
\[Пусть\ x^{2} = t,\ \ t \geq 0,\]
\[4t^{2} - 17t + 4 = 0\]
\[D = 289 - 64 = 225\]
\[t_{1} = \frac{17 + 15}{8} = 4,\]
\[\text{\ \ }t_{2} = \frac{17 - 15}{8} = \frac{2}{8} = \frac{1}{4},\]
\[1)\ x² = 4\]
\[x_{1,2} = \pm 2;\]
\[2)\ x² = \frac{1}{4},\]
\[x_{3,4} = \pm \frac{1}{2}.\]
\[Ответ:x = \pm 2;\ \ x = \pm \frac{1}{2}.\]
\[\textbf{б)}\ 9x^{4} + 77x² - 36 = 0\]
\[Пусть\ x^{2} = t,\ \ t \geq 0,\]
\[9t^{2} + 77t - 36 = 0\]
\[D = 5929 + 1296 = 7225\]
\[t_{1} = \frac{- 77 - 85}{18} = - 9 \Longrightarrow\]
\[\Longrightarrow не\ подходит.\text{\ \ }\]
\[t_{2} = \frac{- 77 + 85}{18} = \frac{4}{9},\]
\[x² = \frac{4}{9},\ \ x_{1,2} = \pm \frac{2}{3}.\]
\[Ответ:x = \pm \frac{2}{3}.\]
\[\textbf{в)}\ 2x^{4} - 9x^{2} - 5 = 0\]
\[Пусть\ x^{2} = t,\ \ t \geq 0,\]
\[2t^{2} - 9t - 5 = 0\]
\[D = 81 + 40 = 121,\]
\[t_{1} = \frac{9 - 11}{4} = - \frac{1}{2} \Longrightarrow\]
\[\Longrightarrow не\ подходит.\]
\[t_{2} = \frac{9 + 11}{4} = 5,\]
\[x² = 5,\ \ x = \pm \sqrt{5}.\]
\[Ответ:x = \pm \sqrt{5}.\]
\[\textbf{г)}\ 6x^{4} - 5x^{2} - 1 = 0\]
\[Пусть\ x^{2} = t,\ \ t \geq 0,\]
\[6t^{2} - 5t - 1 = 0\]
\[D = 25 + 24 = 49\]
\[t_{1} = \frac{5 + 7}{12} = 1,\]
\[t_{2} = \frac{5 - 7}{12} = - \frac{1}{6} \Longrightarrow\]
\[\Longrightarrow не\ подходит.\]
\[x^{2} = 1,\ \ x = \pm 1.\]
\[Ответ:x = \pm 1.\]
