\(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\frac{1^7}{81^7}=\frac{1}{\left(3^4\right)^7}=\frac{1}{3^{28}}>\frac{1}{3^{30}}=\frac{1}{\left(3^5\right)^6}=\frac{1^6}{243^6}=\left(\frac{1}{243}\right)^6\)

=>\(\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)

\(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\frac{1}{3^{28}}\)

\(\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{3^{30}}\)

Vì \(\frac{1}{3^{28}}>\frac{1}{3^{30}}\) nên \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)

Ta có : \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\frac{1}{\left(3 ^4\right)^7}=\frac{1}{3^{28}}\)(1)

        \(\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{^{\left(3^5\right)^6}}=\frac{1}{3^{30}}\left(\frac{1}{243}\right)^6\)

(3/8)⁵= 3⁵/(2³)⁵=243/2¹⁵>243/3¹⁵>125/3¹⁵=5³/(3⁵)³=(5/3⁵)³=(5/243)³

suy ra (3/8)⁵>(5/243)³

a/

+ \(\frac{1}{243^6}=\frac{1}{3^6.81^6}=\frac{1}{3^2.3^4.81^6}=\frac{1}{9.81^7}\) (1)

+ \(80< 81\Rightarrow80^7< 81^7\Rightarrow\frac{1}{80^7}>\frac{1}{81^7}\) (2)

+ \(81^7< 9.81^7\Rightarrow\frac{1}{81^7}>\frac{1}{9.81^7}\) (3)

Từ (1) (2) (3) \(\Rightarrow\frac{1}{80^7}>\frac{1}{243^6}\)

b/ Xem lại đề bài

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