a) Ta có: \(\left(\dfrac{1}{243}\right)^6=\left(\dfrac{1}{3}\right)^{5\cdot6}=\left(\dfrac{1}{3}\right)^{30}\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{28}>\left(\dfrac{1}{243}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{3^4}\right)^7>\left(\dfrac{1}{243}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{81}\right)^7>\left(\dfrac{1}{243}\right)^6\)
mà \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{81}\right)^7\)
nên \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{243}\right)^6\)
\(\left(\dfrac{3}{8}\right)^5\&\left(\dfrac{5}{243}\right)^3\)
\(\left(\dfrac{3}{8}\right)^5=\left(\dfrac{90}{240}\right)^5=\dfrac{90^5}{240^5}\)
\(\left(\dfrac{5}{243}\right)^3=\dfrac{5^3}{243^3}\)
\(=>\dfrac{90^5}{240^5}>\dfrac{5^3}{243^3}\)
\(=>\left(\dfrac{3}{8}\right)^5>\left(\dfrac{5}{243}\right)^3\)
\(\left(\dfrac{1}{80}\right)^7\&\left(\dfrac{1}{243}\right)^6\)
\(\dfrac{1}{80}>\dfrac{1}{81}=\dfrac{1}{3^4}\)
\(=>\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{3^4}\right)^7=\dfrac{1}{3^{7.4}}=\dfrac{1}{3^{28}}>\dfrac{1}{3^{30}}\)
\(=\dfrac{1}{\left(3^5\right)^6}=\left(\dfrac{1}{243}\right)^6\)
\(=>\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{243}\right)^6\)