Iloczyny
: 11 gru 2023, o 11:33
Rozwiázać układ:
\(\displaystyle{ \begin{cases} x_1x_n = 2 \\ x_2(x_n-x_1)=1 \\ ... \\ x_{n-1}(x_n- x_{n-2}) =1 \\ x_n(x_n - x_{n-1}) = 1 \end{cases}}\)
\(\displaystyle{ \begin{cases} x_1x_n = 2 \\ x_2(x_n-x_1)=1 \\ ... \\ x_{n-1}(x_n- x_{n-2}) =1 \\ x_n(x_n - x_{n-1}) = 1 \end{cases}}\)