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a) \(21^{15}\)\(=\left(3.7\right)^{15}\) \(=3^{15}.7^{15}\)
\(27^5.49^8\) \(=\left(3^3\right)^5.\left(7^2\right)^8\)\(=3^{15}.7^{16}\)
Ta thấy \(7^{15}< 7^{16}\)\(\Rightarrow\)\(21^{15}< 27^5.49^8\)
b) \(3^{30}\)\(=3^{2.15}\)\(=\left(3^2\right)^{15}\)\(=9^{15}\)
Ta thấy \(8^{15}< 9^{15}\)
\(\Rightarrow\)\(8^{15}< 3^{30}\)
Ta có: \(21^{15}=\left(3.7\right)^{15}=3^{15}.7^{15}\)
\(27^5.49^8=3^{3.5}.7^{2.8}=3^{15}.7^{16}\)
Vì \(15< 16\)nên \(7^{15}< 7^{16}\Rightarrow3^{15}.7^{15}< 3^{15}.7^{16}\)
Hay \(21^{15}< 27^5.49^8\)
Ta có: \(3^{30}=3^{2.15}=9^{15}\)
Vì \(8< 9\)nên \(8^{15}< 9^{15}\)
Hay \(8^{15}< 3^{30}\)
a)\(\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\)
\(=\left(100+121+144\right)\div\left(169+196\right)\)
\(=365\div365\)
\(=1\)
b) \(1.2.3...9-1.2.3...8-1.2.3...8^2\)
\(=1.2.3...8\left(9-1-8\right)\)
\(=1.2.3...8.0\)
\(=0\)
d) \(1152-\left(374+1152\right)+\left(-65+374\right)\)
\(=1152-374-1152-65+374\)
\(=\left(1152-1152\right)-65+\left(374-374\right)\)
\(=0-65+0\)
\(=-65\)
e) \(13-12+11+10-9+8-7-6+5-4+3+2-1\)
\(=13-\left(12-11\right)+\left(10-9\right)+\left(8-7\right)-\left(6-5\right)-\left(4-3\right)\)\(+\left(2-1\right)\)
\(=13-1+1+1-1-1+1\)
\(=13+0+0+0\)
\(=13\)
\(2.2^2.2^3.2^4.........2^{100}\)
\(=2^{1+2+3+4+......+100}\)
\(=2^{5050}\)