Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1100

\[\boxed{\mathbf{1100}\mathbf{.}}\]

\[1)\ \frac{P_{x + 1}}{P_{x - 1}} = 30\]

\[\frac{(x + 1)!}{(x - 1)!} = 30\]

\[\frac{(x + 1) \bullet x \bullet (x - 1)!}{(x - 1)!} = 30\]

\[x^{2} + x - 30 = 0\]

\[D = 1^{2} + 4 \bullet 30 = 1 + 120 = 121\]

\[x_{1} = \frac{- 1 - 11}{2} = - 6\ \ и\ \ \]

\[x_{2} = \frac{- 1 + 11}{2} = 5.\]

\[Ответ:\ \ x = 5.\]

\[2)\ \frac{P_{x}}{P_{x - 2}} = 42\]

\[\frac{x!}{(x - 2)!} = 42\]

\[\frac{x \bullet (x - 1) \bullet (x - 2)!}{(x - 2)!} = 42\]

\[x^{2} - x - 42 = 0\]

\[D = 1^{2} + 4 \bullet 42 = 1 + 168 = 169\]

\[x_{1} = \frac{1 - 13}{2} = - 6\ \ и\ \ \]

\[x_{2} = \frac{1 + 13}{2} = 7.\]

\[Ответ:\ \ x = 7.\]

\[3)\ \frac{1}{P_{x - 5}} = \frac{56}{P_{x - 3}}\]

\[\frac{1}{(x - 5)!} = \frac{56}{(x - 3)!}\]

\[\frac{(x - 3)!}{(x - 5)!} = 56\]

\[\frac{(x - 3) \bullet (x - 4) \bullet (x - 5)!}{(x - 5)!} = 56\]

\[x^{2} - 4x - 3x + 12 - 56 = 0\]

\[x^{2} - 7x - 44 = 0\]

\[D = 7^{2} + 4 \bullet 44 = 49 + 176 =\]

\[= 225\]

\[x_{1} = \frac{7 - 15}{2} = - 4\ \ и\ \ \]

\[x_{2} = \frac{7 + 15}{2} = 11.\]

\[Ответ:\ \ x = 11.\]

\[4)\ \frac{1}{P_{x - 4}} = \frac{110}{P_{x - 2}}\]

\[\frac{1}{(x - 4)!} = \frac{110}{(x - 2)!}\]

\[\frac{(x - 2)!}{(x - 4)!} = 110\]

\[\frac{(x - 2) \bullet (x - 3) \bullet (x - 4)!}{(x - 4)!} = 110\]

\[x^{2} - 3x - 2x + 6 - 110 = 0\]

\[x^{2} - 5x - 104 = 0\]

\[D = 5^{2} + 4 \bullet 104 = 25 + 416 =\]

\[= 441\]

\[x_{1} = \frac{5 - 21}{2} = - 8\ \ и\ \ \]

\[x_{2} = \frac{5 + 21}{2} = 13.\]

\[Ответ:\ \ x = 13.\]

\[5)\ A_{x + 1}^{3} = 72(x - 1)\]

\[\frac{(x + 1)!}{(x + 1 - 3)!} = 72(x - 1)\]

\[\frac{(x + 1) \bullet x \bullet (x - 1) \bullet (x - 2)!}{(x - 2)!} =\]

\[= 72(x - 1)\]

\[(x + 1) \bullet x = 72\]

\[x^{2} + x - 72 = 0\]

\[D = 1^{2} + 4 \bullet 72 = 1 + 288 = 289\]

\[x_{1} = \frac{- 1 - 17}{2} = - 9\ \ и\ \]

\[x_{2} = \frac{- 1 + 17}{2} = 8.\]

\[Ответ:\ \ x = 8.\]

\[6)\ A_{x - 1}^{4} = 40(x - 2)(x - 3)\]

\[\frac{(x - 1)!}{(x - 1 - 4)!} = 40(x - 2)(x - 3)\]

\[\frac{(x - 1)!}{(x - 5)!} = 40(x - 2)(x - 3)\]

\[(x - 1)(x - 4) = 40\]

\[x^{2} - 4x - x + 4 - 40 = 0\]

\[x^{2} - 5x - 36 = 0\]

\[D^{2} = 5^{2} + 36 \bullet 4 = 25 + 144 =\]

\[= 169\]

\[x_{1} = \frac{5 - 13}{2} = - 4\ \ и\ \ \]

\[x_{2} = \frac{5 + 13}{2} = 9.\]

\[Ответ:\ \ x = 9.\]

\[7)\ 5C_{n + 1}^{3} = 8C_{n}^{4}\]

\[5 \bullet \frac{(n + 1)!}{(n + 1 - 3)! \bullet 3!} =\]

\[= 8 \bullet \frac{n!}{(n - 4)! \bullet 4!}\]

\[\frac{5 \bullet (n + 1) \bullet n \bullet (n - 1)(n - 2)!}{(n - 2)! \bullet 3 \bullet 2} =\]

\[\frac{5 \bullet (n + 1)}{6} = \frac{(n - 2)(n - 3)}{3}\]

\[5(n + 1) = 2(n - 2)(n - 3)\]

\[5n + 5 = 2n^{2} - 6n - 4n + 12\]

\[2n^{2} - 15n + 7 = 0\]

\[D = 15^{2} - 4 \bullet 2 \bullet 7 = 225 - 56 =\]

\[= 169\]

\[x_{1} = \frac{15 - 13}{2 \bullet 2} = \frac{2}{4} = \frac{1}{2}\text{\ \ }и\ \ \]

\[x_{2} = \frac{15 + 13}{2 \bullet 2} = \frac{28}{4} = 7.\]

\[Ответ:\ \ x = 7.\]

\[8)\ C_{n}^{3} = 4C_{n - 2}^{2}\]

\[\frac{n!}{(n - 3)! \bullet 3!} = 4 \bullet \frac{(n - 2)!}{(n - 2 - 2)! \bullet 2!}\]

\[\frac{n \bullet (n - 1) \bullet (n - 2) \bullet (n - 3)!}{(n - 3)! \bullet 3 \bullet 2} =\]

\[= \frac{4 \bullet (n - 2) \bullet (n - 3) \bullet (n - 4)!}{(n - 4)! \bullet 2}\]

\[\frac{n(n - 1)}{6} = 2(n - 3)\]

\[n(n - 1) = 12(n - 3)\]

\[n^{2} - n = 12n - 36\]

\[n^{2} - 13n + 36 = 0\]

\[D = 13^{2} - 4 \bullet 36 = 169 - 144 =\]

\[= 25\]

\[x_{1} = \frac{13 - 5}{2} = 4\ \ и\ \]

\[x_{2} = \frac{13 + 5}{2} = 9.\]

\[Ответ:\ \ n_{1} = 9;\ \ n_{2} = 4.\]