`(1/243)^9 = [1/(3^5)]^9 = [(1/3)^5]^9=(1/3)^13`

Vì: `1/3 > 1/83`

`=> (1/3)^13 > 1/(83)^13`.

1\243=1\(81.3) 
(1\243)^9=1\((81^9.3^9)=1\(81^9.27^3)> 1\(81^9.81^3) >1\(83^12)>1\(83^13)

k cho mình nhé

a) Vì \(\dfrac{1}{24}< \dfrac{1}{83}\) 

⇒ \(\dfrac{1}{24^9}>\dfrac{1}{83^{13}}\)

a) \(\left(\dfrac{1}{24}\right)^9>\left(\dfrac{1}{27}\right)^9=\dfrac{1}{3^{27}}\)

\(\left(\dfrac{1}{83}\right)^{13}< \left(\dfrac{1}{81}\right)^{13}=\dfrac{1}{3^{52}}\)

Mà \(\dfrac{1}{3^{27}}>\dfrac{1}{3^{52}}\)

\(\Rightarrow\left(\dfrac{1}{24}\right)^9>\left(\dfrac{1}{83}\right)^{13}\)

b) \(3^{300}=\left(3^3\right)^{100}=27^{100}\)

\(5^{199}< 5^{200}=\left(5^2\right)^{100}=25^{100}\)

Mà \(25^{100}< 27^{100}\)

\(\Rightarrow5^{199}< 3^{300}\)

\(\Rightarrow\dfrac{1}{5^{199}}>\dfrac{1}{3^{300}}\)

\(a,\left(\dfrac{1}{24}\right)^9=\dfrac{1}{24^9};\left(\dfrac{1}{83}\right)^{13}=\dfrac{1}{83^{13}};24^9< 83^{13}\left(24< 83;9< 13\right)\\ \Rightarrow\dfrac{1}{24^9}< \dfrac{1}{83^{13}}\Rightarrow\left(\dfrac{1}{24}\right)^9< \left(\dfrac{1}{83}\right)^{13}\\ b,3^{300}=27^{100}>25^{100}=5^{200}>5^{199}\\ \Rightarrow\dfrac{1}{3^{300}}< \dfrac{1}{5^{199}}\)

\(\left(\frac{1}{243}\right)^9=\frac{1}{243^9}=\frac{1}{\left(3^5\right)^9}=\frac{1}{3^{45}}\)

\(\left(\frac{1}{83}\right)^{13}< \left(\frac{1}{81}\right)^{13}=\frac{1}{81^{13}}=\frac{1}{\left(3^4\right)^{13}}=\frac{1}{3^{52}}\)

Có \(3^{45}< 3^{52}\Rightarrow\frac{1}{3^{45}}>\frac{1}{3^{52}}\)

suy ra \(\left(\frac{1}{243}\right)^9>\left(\frac{1}{83}\right)^{13}\).

Ta có : 

\(\frac{1}{243^9}=\frac{1}{\left(81.3\right)^9}=\frac{1}{81^9.27^3}>\frac{1}{81^9.81^3}=\frac{1}{81^{11}}>\frac{1}{8^{12}}>\frac{1}{8^{13}}\)

\(\Rightarrow\frac{1}{243^9}>\frac{1}{8^{13}}\)

Ta có :

\(\frac{1}{243^9}=\frac{1}{\left(81.3\right)^9}=\frac{1}{81^9.27^3}>\frac{1}{81^9.81^3}=\frac{1}{81^{11}}>\frac{1}{8^{12}}>\frac{1}{8^{13}}\)

\(\Rightarrow\frac{1}{243^9}>\frac{1}{83^{13}}\)

mình chắc chắn luôn

-https://olm.vn/hoi-dap/detail/77727486175.html