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

\(\left(\dfrac{3}{8}\right)^5\&\left(\dfrac{5}{243}\right)^3\)
\(\left(\dfrac{3}{8}\right)^5=\left(\dfrac{90}{240}\right)^5=\dfrac{90^5}{240^5}\)

\(\left(\dfrac{5}{243}\right)^3=\dfrac{5^3}{243^3}\)

\(=>\dfrac{90^5}{240^5}>\dfrac{5^3}{243^3}\)

\(=>\left(\dfrac{3}{8}\right)^5>\left(\dfrac{5}{243}\right)^3\)

\(\left(\dfrac{1}{80}\right)^7\&\left(\dfrac{1}{243}\right)^6\)

\(\dfrac{1}{80}>\dfrac{1}{81}=\dfrac{1}{3^4}\)

\(=>\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{3^4}\right)^7=\dfrac{1}{3^{7.4}}=\dfrac{1}{3^{28}}>\dfrac{1}{3^{30}}\)

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

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

giúp minh

\(7^{80}=\left(7^4\right)^{20}=2401^{20};5^{100}=\left(5^5\right)^{20}=3125^{20}\\ Vì:2401^{20}< 3125^{20}\left(Do:2401< 3125\right).Nên:7^{80}< 5^{100}\)

chơi nhau à

có cc=)))

Giải hộ đi mà =//

em trả lời ccaua này hi vọng thầy còn nhớ em

a) -9/4<`1/3

c) 9/-5<7/-10

a) \(\dfrac{-9}{4}< 0\)

\(0< \dfrac{1}{3}\)

Do đó: \(\dfrac{-9}{4}< \dfrac{1}{3}\)

a) Ta có: 3,23 < 3,32 nên -3,23 > -3,32

b) Ta có: \( - \frac{7}{3} = \frac{{ - 28}}{{12}}; - 1,25 = \frac{{ - 125}}{{100}} = \frac{{ - 5}}{4} = \frac{{ - 15}}{{12}}\)

Vì -28 < -15 nên \(\frac{{ - 28}}{{12}} < \frac{{ - 15}}{{12}}\) hay \( - \frac{7}{3}\) < -1,25

Bài 1

a: 11/12=1-1/12

23/24=1-1/24

mà -1/12>-1/24

nên 11/12>23/24

b: -3/20=-9/60

-7/12=-35/60

mà -9>-35

nên -3/20>-7/12