Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1064

\[\boxed{\mathbf{1064}\mathbf{.}}\]

\[1)\ 6! \bullet 7 = 1 \bullet 2 \bullet 3 \bullet 4 \bullet 5 \bullet 6 \bullet 7 =\]

\[= 7!;\]

\[5)\ k! \bullet (k + 1) = (k + 1)!;\]

\[6)\ (k - 1)! \bullet k = k!;\]

\[7)\ (k - 1)! \bullet k \bullet (k + 1) =\]

\[= (k + 1)!;\]

\[8)\ (k - 2)! \bullet (k - 1) \bullet k = k!;\]

\[9)\ (k - 4)! \bullet \left( k^{2} - 5k + 6 \right)\]

\[k^{2} - 5k + 6 = 0\]

\[D = 5^{2} - 4 \bullet 6 = 25 - 24 = 1\]

\[k_{1} = \frac{5 - 1}{2} = 2;\text{\ \ }k_{2} = \frac{5 + 1}{2} = 3;\]

\[k^{2} - 5k + 6 = (k - 3)(k - 2).\]

\[(k - 4)! \bullet (k - 3) \bullet (k - 2) =\]

\[= (k - 2)!\]

\[10)\ (k - 3)! \bullet \left( k^{2} - 3k + 2 \right)\]

\[k^{2} - 3k + 2 = 0\]

\[D = 3^{2} - 4 \bullet 2 = 9 - 8 = 1\]

\[k_{1} = \frac{3 - 1}{2} = 1;\text{\ \ }k_{2} = \frac{3 + 1}{2} = 2;\]

\[k^{2} - 3k + 2 = (k - 2)(k - 1).\]

\[(k - 3)! \bullet (k - 2) \bullet (k - 1) =\]

\[= (k - 1)!\]