15 mũ 12 và 81mũ 3 ,125 mũ5
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Ta có :
\(15^{12}=\left(3.5\right)^{12}=3^{12}.5^{12}\)
\(81^3.125^5=\left(3^4\right)^3.\left(5^3\right)^5=3^{12}.5^{15}\)
Mà \(3^{12}.5^{15}>3^{12}.5^{12}\)
\(\Rightarrow15^{12}< 81^3.125^5\)
a,
15^12=(3*5)^12=3^12*5^12
81^3*125^5=(3^4)^3*(5^3)^5=3^12*5^15
Vì 12<15 suy ra 5^12<5^15
Suy ra 3^12*5^12<3^12*5^15
\(a.81^3.125^5=\left(3^4\right)^3.\left(5^3\right)^5=3^{12}.5^{15}=3^{12}.5^{12}.5^3=\left(3.5\right)^{12}.5^3=15^{12}.5^3>15^{12}\)
\(b.4^{20}.81^{12}=\left(2^2\right)^{20}.\left(9^2\right)^{12}=2^{40}.9^{24}=2^{20}.2^{20}.9^{20}.9^4=\left(2.9\right)^{20}.2^{20}.9^4=18^{20}.2^{20}.9^4>18^{20}\)
\(c.73^{75}=\left(73^3\right)^{25}=389017^{25}\)
\(107^{50}=107^{2.50}=\left(107^2\right)^{25}=11449^{25}\)
Vì \(389017^{25}>11449^{25}\Rightarrow73^{75}>107^{50}\)
\(5^{12}.7-5^{11}.10\)
\(=5^{11}.\left(5.7-10\right)\)
\(=5^{11}.25\)
\(=5^{11}.5^2\)
\(=5^{13}\)
\(2^{20}.15+2^{20}.85\)
\(=2^{20}.5\left(3+17\right)\)
\(=2^{20}.100\)
\(=104857600\)
\(125^3:25^4\)
\(=\left(5^3\right)^3:\left(5^2\right)^4\)
\(=5^9:5^8\)
\(=5\)
\(24^4:3^4-32^{12}:16^{12}\)
\(=\left(24:3\right)^4-\left(32:16\right)^{12}\)
\(=8^4-2^{12}\)
\(=0\)
\(a,4^{333}=\left(4^3\right)^{111}=64^{111}< 81^{111}=\left(3^4\right)^{111}=3^{444}< 3^{445}\\ b,39^{15}>36^{15}=\left(6^2\right)^{15}=6^{30}\\ c,25^{45}=\left(5^2\right)^{45}=5^{90}< 5^{102}=\left(5^3\right)^{34}=125^{34}\)