\(\left\{{}\begin{matrix}99^{29}=\left(99^2\right)^{10}=9801^{10}\\9999^{10}=9999^{10}\end{matrix}\right.\)
\(9801^{10}< 9999^{10}\Leftrightarrow99^{20}< 9999^{10}\)
\(\left\{{}\begin{matrix}\left|-\dfrac{25}{46}\right|=\dfrac{25}{46}>0\\\left(-\dfrac{25}{46}\right)^{2005}< 0\end{matrix}\right.\)
\(\Rightarrow\left(-\dfrac{25}{46}\right)^{2005}< \left|-\dfrac{25}{46}\right|\)
a/ Ta có :
\(99^{20}=\left(99^2\right)^{10}=9081^{10}\)
Vì \(9081^{10}< 9999^{10}\Leftrightarrow99^{20}< 9999^{10}\)
b/ Ta có :
\(\left|\dfrac{-25}{46}\right|=\dfrac{25}{46}>0\)
\(\left(\dfrac{-25}{46}\right)^{2005}< 0\)
\(\Leftrightarrow\left|\dfrac{-25}{46}\right|>\left(\dfrac{-25}{46}\right)^{2005}\)
a) Ta có: \(99^{20}=\left(99^2\right)^{10}=\left(99.99\right)^{10}\)
\(9999^{10}=\left(99.101\right)^{10}\)
Rõ ràng ta thấy:
\(99.99< 99.101\)
Nên \(\left(99.99\right)^{10}< \left(99.101\right)^{10}\)
Vậy \(99^{20}< 9999^{10}\)
b) Ta có: \(\left|-\dfrac{25}{46}\right|=\dfrac{25}{46}>0\)
\(\left(-\dfrac{25}{46}\right)^{2005}< 0\) (theo tính chất số mũ của số hữu tỉ)
Vậy \(\left|-\dfrac{25}{46}\right|>\left(-\dfrac{25}{46}\right)^{2005}\)
