b) Ta có:
\(64^8=\left(8^2\right)^8=8^{16}=\left(2^3\right)^{16}=2^{48}\)
\(16^{12}=\left(2^4\right)^{12}=2^{48}\)
Vì \(2^{48}=2^{48}\) nên \(64^8=16^{12}\)
a) Ta có:
\(10^{20}=\left(2.5\right)^{20}=2^{20}.5^{20}\)
\(90^{10}=\left(2^2.3.5\right)^{10}=2^{20}.3^{10}.5^{10}\)
Vì \(2^{20}.5^{20}< 2^{20}.3^{10}.5^{10}\) nên \(10^{20}>90^{10}\)
tick nha:
a) 10\(^{20}\) = (10\(^2\))\(^{10}\) = 100\(^{10}\) > 90\(^{10}\)
Vậy 10\(^{20}\) > 90\(^{10}\) .
b) 64\(^8\) = (2\(^6\))\(^8\) = 2\(^{48}\).
16\(^{12}\) = (2\(^4\))\(^{12}\) = 2\(^{48}\).
Vậy 64\(^8\) = 16\(^{12}\) .
\(a.\)
\(Ta\) \(c\text{ó}:\)
\(10^{20}=\left(2.5\right)^{20}=2^{20}.5^{20}\)
\(90^{10}=\left(2^2.3.5\right)^{10}=2^{20}.3^{10}.5^{10}\)
Vì \(2^{20}.5^{20}< 2^{20}.3^{10}.5^{10}\) nên \(10^{20}>90^{10}\)
Vậy : \(10^{20}>90^{10}\)
\(b.\)
\(Ta\) \(c\text{ó}:\)
\(64^8=\left(8^2\right)^8=8^{16}=\left(2^3\right)^{16}=2^{48}\)
\(16^{12}=\left(2^4\right)^{12}=2^{48}\)
Vì \(2^{48}=2^{48}\) nên \(64^8=16^{12}\)
Vậy : \(64^8=16^{12}\)