номер 111 гдз 8 класс алгебра Муравин Муравин

1)\begin{equation}A_{n}^{2}+C_{n}^{1}=256 \end{equation} \begin{equation}A_{n}^{2}=\frac{n!}{\left ( n-2 \right )!}=\frac{\left ( n-2 \right )\left ( n-1 \right )\cdot n}{n-2}= \end{equation} \begin{equation}=n\left ( n-1 \right ) \end{equation} \begin{equation}C_{n}^{1}=\frac{n!}{\left ( n-1 \right )!}=\frac{\left ( n-1 \right )\cdot n}{n-1}=n \end{equation}
n + n(n - 1)=256
n + n² - n = 256
n² = 256
n = 16
Ответ: n = 16
2)\begin{equation}\frac{C_{2n+1}^{n-1}}{C_{2n}^{n+1}}=\frac{13}{7} \end{equation} \begin{equation}C_{2n+1}^{n-1}=\frac{2n+1}{\left ( n-1 \right )\left ( 2n+1-\left ( n-1 \right ) \right )}= \end{equation} \begin{equation}=\frac{2n+1}{\left ( n-1 \right )\left ( 2n+1-n+1 \right )}=\frac{2n+1}{\left ( n-1 \right )\left ( n+2 \right )} \end{equation} \begin{equation}C_{2n}^{n+1}=\frac{2n}{\left ( n+1 \right )\left ( 2n-\left ( n+1 \right ) \right )}= \end{equation} \begin{equation}=\frac{2n}{\left ( n+1 \right )\left ( n-1 \right )} \end{equation} \begin{equation}\frac{2n+1}{\left ( n+2 \right )\left ( n-1 \right )}:\frac{2n}{\left ( n+1 \right )\left ( n-1 \right )}= \end{equation} \begin{equation}=\frac{2n+1}{\left ( n+2 \right )\left ( n-1 \right )}\cdot \frac{\left ( n+1 \right )\left ( n-1 \right )}{2n}= \end{equation} \begin{equation}=\frac{\left ( 2n+1 \right )\left ( n+1 \right )}{\left ( n+2 \right )\cdot 2n} \end{equation} \begin{equation}\frac{\left ( 2n+1 \right )\left ( n+1 \right )}{\left ( n+2 \right )2n}=\frac{13}{7} \end{equation}
7(2n + 1)(n + 1) = 13 · 2n(n + 2)
7(2n² + 2n + n + 1) = 26n(n + 2)
14n² + 21n + 7 = 26n² + 52n
26n² - 14n² - 21n + 52n - 7 = 0
12n² + 31n - 7 = 0
D = 31² - 4 · 12 · (-7) = 961 + 336 = 1297 \begin{equation}\sqrt{D}=36 \end{equation} \begin{equation}n=\frac{-31\pm 36}{24}=-3;0,2 \end{equation} 3)\begin{equation}A_{n+1}^{2}+C_{n+1}^{n-1}=18\left ( n+1 \right ) \end{equation} \begin{equation}A_{n+1}^{2}=\frac{\left ( n+1 \right )!}{\left ( n+1-2 \right )!}=\frac{n+1}{n-1} \end{equation} \begin{equation}C_{n+1}^{n-1}=\frac{\left ( n+1 \right )!}{\left ( n-1 \right )\left ( n+1-\left ( n-1 \right ) \right )}= \end{equation} \begin{equation}=\frac{n+1}{\left ( n-1 \right )\left ( n+1-n+1 \right )}=\frac{n+1}{2\cdot \left ( n-1 \right )} \end{equation} \begin{equation}\frac{n+1}{n-1}+\frac{n+1}{2\left ( n-1 \right )}=18\left ( n+1 \right ) \end{equation}
2(n + 1) + n + 1 = 36(n² - 1)
2n + 2 + n + 1 = 36n² - 36
3n + 3 = 36(n² - 1)
3(n + 1) = 36(n² - 1) : 3

n + 1 = 12(n² - 1)
n + 1 = 12n² - 12
12n² - n - 1 - 12 = 0
12n² - n - 13 = 0
D = (-1)² - (-13) · 4 · 12 = 1 + 624 = 625 \begin{equation}n_{1;2}=\frac{1\pm 25}{24}=-\frac{24}{24}=-1; \end{equation} \begin{equation}\frac{26}{24}=\frac{13}{12} \end{equation}
или
n + 1 = 12(n + 1)(n - 1) : n + 1
1 = 12(n - 1)
1 = 12n - 12
12n = 13 \begin{equation}n=\frac{13}{12} \end{equation} 4)\begin{equation}A_{n}^{3}+C_{n}^{n-2}=7,5n \end{equation} \begin{equation}A_{n}^{3}=\frac{n!}{\left ( n-3 \right )!}=\frac{n}{n-3} \end{equation} \begin{equation}C_{n}^{n-2}=\frac{n!}{\left ( n-\left ( n-2 \right ) \right )!}=\frac{n}{n-n+2}=\frac{n}{2} \end{equation} \begin{equation}\frac{n}{n-3}+\frac{n}{2}=7,5n \end{equation}
2n + n² - 3n = 15n(n - 3)
n² - n = 15n² - 45n
15n² - 45n + n - n² = 0
14n² - 44n = 0
n(14n - 44) = 0

n = 0

14n - 44 = 0
14n = 44 \begin{equation}n=\frac{44}{14}=3 \end{equation}