51 = 5
5 > 1
70 = 1
1 = 1
2-3 = \(\frac{1}{2^3}\) = \(\frac{1}{8}\) = 0.125
0.125 < 1
5 ⋅ 5-2 = 5 ⋅ \(\frac{1}{5^2}\) = \(\frac{5}{25}\) = \(\frac{1}{5}\) = 0.2
0.2 < 1
30 ⋅ 7-3 = 1 ⋅ \(\frac{1}{7^3}\) = \(\frac{1}{343}\) ≈ 0.0029
0.0029 < 1
\(\frac{2^7 ⋅ 2^3}{2^{13}}\) = \(\frac{2^{10}}{2^{13}}\) = 2-3 = \(\frac{1}{2^3}\) = \(\frac{1}{8}\) = 0.125
0.125 < 1
27 ⋅ 2-5 = 22 = 4
4 > 1
70 ⋅ 82 ⋅ 2-3 = 1 ⋅ 64 ⋅ \(\frac{1}{8}\) = 8
8 > 1
\(\frac{3^5 ⋅ 5^5}{15^5}\) = \(\frac{15^5}{15^5}\) = 1
1 = 1
6-1 = \(\frac{1}{6}\) ≈ 0.167
0.167 < 1
2-2 ⋅ 3-2 = \(\frac{1}{2^2}\) ⋅ \(\frac{1}{3^2}\) = \(\frac{1}{4}\) ⋅ \(\frac{1}{9}\) = \(\frac{1}{36}\) ≈ 0.028
0.028 < 1
(23)2 ⋅ 2-7 = 26 ⋅ 2-7 = 2-1 = \(\frac{1}{2}\) = 0.5
0.5 < 1
c-9 = \(\frac{1}{c^9}\)
Если c > 1, то c-9 < 1; если 0 < c < 1, то c-9 > 1; если c = 1, то c-9 = 1.
2-5 = \(\frac{1}{2^5}\) = \(\frac{1}{32}\) = 0.03125
0.03125 < 1
k0 = 1
1 = 1
Ответ: См. решение выше