Ta có :

 \(\frac{1^{500}}{2}=\frac{1}{2}\)

\(\frac{1^{300}}{3}=\frac{1}{3}\)

Mà 3>2

\(\Rightarrow\frac{1}{2}>\frac{1}{3}\)

Hay \(\frac{1^{500}}{2}>\frac{1^{300}}{3}\)

Ta có:

(-1/5)³⁰⁰ = (-1)³⁰⁰/5³⁰⁰ = 1/(5³)¹⁰⁰ = 1/125¹⁰⁰

(-1/3)⁵⁰⁰ = (-1)⁵⁰⁰/3⁵⁰⁰ = 1/(3⁵)¹⁰⁰ = 1/243¹⁰⁰

Vì 125¹⁰⁰ < 243¹⁰⁰

=> 1/125¹⁰⁰ > 1/243¹⁰⁰

=> (-1/5)³⁰⁰ > (-1/3)⁵⁰⁰

Ta có : \(\left(-\frac{1}{5}\right)^{300}=\left(-\frac{1}{5}\right)^{3.100}=\left(-\frac{1}{125}\right)^{100}=\left(\frac{1}{125}\right)^{100}\)

            \(\left(-\frac{1}{3}\right)^{500}=\left(-\frac{1}{3}\right)^{5.100}=\left(-\frac{1}{243}\right)^{100}=\left(\frac{1}{243}\right)^{100}\)

Mà \(125< 243\Rightarrow\frac{1}{125}>\frac{1}{243}\Rightarrow\left(\frac{1}{125}\right)^{100}>\left(\frac{1}{243}\right)^{100}\)

\(=>\left(-\frac{1}{5}\right)^{300}>\left(-\frac{1}{3}\right)^{500}\)

Ta có:

(-1/5)³⁰⁰ = (-1)³⁰⁰/5³⁰⁰ = 1/(5³)¹⁰⁰ = 1/125¹⁰⁰

(-1/3)⁵⁰⁰ = (-1)⁵⁰⁰/3⁵⁰⁰ = 1/(3⁵)¹⁰⁰ = 1/243¹⁰⁰

Vì 125¹⁰⁰ < 243¹⁰⁰

=> 1/125¹⁰⁰ > 1/243¹⁰⁰

=> (-1/5)³⁰⁰ > (-1/3)⁵⁰⁰

\(\frac{-1^{100}}{16^{100}}=\frac{-1}{\left(2^4\right)^{100}}=\frac{-1}{2^{400}};\frac{-1^{500}}{2^{500}}=\frac{-1}{2^{500}}\)

Vì 2⁴⁰⁰<2⁵⁰⁰ => \(\frac{-1}{2^{400}}>\frac{-1}{2^{500}}\)=>\(\frac{-1^{100}}{16^{100}}>\frac{-1^{500}}{2^{500}}\)

không

em không thể trả lời được

cho em nhé 

kết bạn với em nhé

a) Ta có :\(\left(\frac{-1}{5}\right)^{300}=\frac{-1^{300}}{5^{300}}=\frac{1}{125^{100}}\)

\(\left(\frac{-1}{3}\right)^{500}=\frac{-1^{500}}{3^{500}}=\frac{1}{243^{100}}\)

Mà \(\frac{1}{125^{100}}>\frac{1}{243^{100}}\)

\(\Rightarrow\left(\frac{-1}{5}\right)^{300}>\left(\frac{-1}{3}\right)^{500}\)

b)Ta có :\(2^{90}=\left(2^{15}\right)^6=32768^6\)

\(5^{36}=\left(5^6\right)^6=15625^6\)

Vì \(32768^6>15625^6\Rightarrow2^{90}>5^{36}\)

a.Ta có: \(\left(\frac{-1}{5}\right)^{300}=\left(\frac{-1}{5}^3\right)^{100}=\left(\frac{-1}{125}\right)^{100}=\left(\frac{1}{125}\right)^{100}\)

\(\left(\frac{-1}{3}\right)^{500}=\left(\frac{-1}{3}^5\right)^{100}=\left(\frac{-1}{243}\right)^{100}=\left(\frac{1}{234}\right)^{100}\)

Mà: \(\frac{1}{125}>\frac{1}{234}\Rightarrow\left(\frac{1}{125}\right)^{100}>\left(\frac{1}{234}\right)^{100}\)

Vậy \(\left(\frac{-1}{5}\right)^{300}>\left(\frac{-1}{3}\right)^{500}\)

b.Ta có: \(2^{90}=\left(2^{10}\right)^9=1024^9\)

\(5^{36}=\left(5^4\right)^9=625^9\)

Mặt khác: \(1024>625\Rightarrow1024^9>625^9\)

 Vậy \(2^{90}>5^{36}\)

ta có: \(\frac{300}{-299}< -1\)

\(\frac{-500}{507}>-1\)

\(\Rightarrow\frac{300}{-299}< \frac{-500}{507}\Rightarrow x< y\)

\(x=\frac{300}{-299};y=\frac{-500}{507}\)

Ta có ;  \(\frac{300}{299}>1\)             \(\frac{500}{507}< 1\)

\(\frac{300}{299}>\frac{500}{507}\)

\(\Rightarrow\frac{300}{-299}< \frac{-500}{507}\)

hok tốt

\(^{x=\frac{300}{-299}}\)<\(^{x=\frac{-500}{507}}\)

k mik nha!

Bài 1: \(\left(\frac{-1}{16}\right)^{100}=\frac{1}{\left(2^4\right)^{100}}=\frac{1}{2^{400}}>\frac{1}{2^{500}}=\left(\frac{-1}{2}\right)^{500}.\)

Bài 2: \(100^{99}+1>100^{68}+1\Rightarrow\frac{1}{100^{99}+1}< \frac{1}{100^{68}+1}\Rightarrow\frac{-99}{100^{99}+1}>\frac{-99}{100^{68}+1}\)

\(\Rightarrow100+\frac{-99}{100^{99}+1}>100+\frac{-99}{100^{68}+1}\Rightarrow\frac{100^{100}+1}{100^{99}+1}>\frac{100^{69}+1}{100^{68}+1}\)

\(\left(\frac{1}{3}\right)^{500}=\left(\frac{1}{3}^5\right)^{100}=\frac{1}{243}^{100}\)

\(\left(\frac{1}{5}\right)^{300}=\left(\frac{1}{5}^3\right)^{100}=\frac{1}{125}^{100}\)

Vì \(\frac{1}{243}<\frac{1}{125}=>\frac{1}{243}^{100}<\frac{1}{125}^{100}=>\left(\frac{1}{3}\right)^{500}<\left(\frac{1}{5}\right)^{300}\)

3^-500=(3⁵)^-100= 243^-100

5^-300= (5³)^-100 =125^-100

^{243>125 =>}    243^-100<125^-100

Hay 3^-500 <5^-300

x<y nhé bạn :)