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\(\left(-\frac{1}{5}\right)^{300}=-\frac{1^{300}}{5^{300}}=-\frac{1}{5^{300}}\)
\(\left(-\frac{1}{5}\right)^{500}=-\frac{1^{500}}{5^{500}}=-\frac{1}{5^{500}}\)
Ta có :
\(5^{300}< 5^{500}\)
\(\Rightarrow-5^{300}>-5^{500}\)
\(\Rightarrow-\frac{1}{5^{300}}>-\frac{1}{5^{500}}\)
\(\Rightarrow\left(-\frac{1}{5}\right)^{300}>\left(-\frac{1}{5}\right)^{500}\)
a/
\(\left(-\frac{1}{16}\right)^{1000}=\left(-\frac{1}{2^4}\right)^{1000}=\left(-\frac{1}{2}\right)^{4000}.\)
Do \(\left(\frac{1}{2}\right)^{4000}>\left(\frac{1}{2}\right)^{5000}\Rightarrow\left(-\frac{1}{2}\right)^{4000}< \left(-\frac{1}{2}\right)^{5000}\Rightarrow\left(-\frac{1}{16}\right)^{1000}< \left(-\frac{1}{2}\right)^{5000}\)
b/
\(3^{400}=\left(3^4\right)^{100}=81^{100}\)
\(4^{300}=\left(4^3\right)^{100}=64^{100}\)
\(\Rightarrow81^{100}>64^{100}\Rightarrow3^{400}>4^{300}\)
Ta có: \(5^{300}=\left(5^3\right)^{100}=125^{100}\)(1)
\(3^{500}=\left(3^5\right)^{100}=243^{100}\)(2)
Từ (1) và (2) suy ra \(5^{300}< 3^{500}\)
\(\Rightarrow\frac{1}{5^{300}}>\frac{1}{3^{500}}\)
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 :
\(-\left(\frac{1}{3}\right)^{500}=-\left(\frac{1}{3}\right)^{5.100}=\) \(-\left(\frac{1}{243}\right)^{100}\)
\(-\left(\frac{1}{5}\right)^{300}=-\left(\frac{1}{5}\right)^{3.100}\) =\(-\left(\frac{1}{125}\right)^{100}\)
Vì \(-\left(\frac{1}{125}\right)< -\left(\frac{1}{243}\right)\)nên \(-\left(\frac{1}{3}\right)^{500}>-\left(\frac{1}{5}\right)^{300}\)