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gọi \(A=\frac{2015^{2015}+1}{2015^{2016}+1};B=\frac{2015^{2014}+1}{2015^{2015}+1}\)
\(\Rightarrow A=\frac{2015^{2015}+1}{2015^{2016}+1}<\frac{2015^{2015}+2014+1}{2015^{2016}+2014+1}=\frac{2015^{2015}+2015}{2015^{2016}+2015}=\frac{2015\left(2015^{2014}+1\right)}{2015\left(2015^{2015}+1\right)}=\frac{2015^{2014}+1}{2015^{2015}+1}=B\)
\(1.\)
a, \(27^{265}\)và \(81^{199}\)
\(27^{265}=\left(3^3\right)^{265}=3^{795}\)
\(81^{199}=\left(3^4\right)^{199}=3^{796}\)
\(\Rightarrow3^{795}< 3^{796}hay27^{265}< 81^{199}\)
b, \(1024^{15}=\left(2^{10}\right)^{15}=2^{150}\)
\(128^{21}=\left(2^7\right)^{21}=2^{147}\)
\(2^{150}>2^{147}.hay.1024^{15}>128^{21}\)
b) Ta có: \(5-\left|3x-1\right|=3\)
\(\Leftrightarrow\left|3x-1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=-2\\3x-1=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=1\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{1}{3};1\right\}\)
c) Ta có: \(\left(1-2x\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2\\2x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;2\right\}\)