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Ta có :
\(\frac{1}{243^9}=\frac{1}{\left(81.3\right)^9}=\frac{1}{81^9.27^3}>\frac{1}{81^9.81^3}=\frac{1}{81^{11}}>\frac{1}{8^{12}}>\frac{1}{8^{13}}\)
\(\Rightarrow\frac{1}{243^9}>\frac{1}{8^{13}}\)
a) Ta có :
\(27^{27}>27^{26}=\left(27^2\right)^{13}=729^{13}>243^{13}\)
\(\Rightarrow27^{27}>243^{13}\)
\(\Rightarrow-27^{27}< -243^{13}\)
\(\Rightarrow\left(-27\right)^{27}< \left(-243\right)^{13}\)
b) \(\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{8}\right)^{26}=\left(\dfrac{1}{8^2}\right)^{13}=\left(\dfrac{1}{64}\right)^{13}>\left(\dfrac{1}{128}\right)^{13}\)
\(\Rightarrow\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{128}\right)^{13}\)
\(\Rightarrow\left(-\dfrac{1}{8}\right)^{25}< \left(-\dfrac{1}{128}\right)^{13}\)
c) \(4^{50}=\left(4^5\right)^{10}=1024^{10}\)
\(8^{30}=\left(8^3\right)^{10}=512^{10}< 1024^{10}\)
\(\Rightarrow4^{50}>8^{30}\)
d) \(\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{9}\right)^{12}< \left(\dfrac{1}{27}\right)^{12}\)
\(\Rightarrow\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{27}\right)^{12}\)
\(\left(-27\right)^{27}=\left(-3\right)^{3^{27}}=\left(-3\right)^{81}\)
\(\left(-243\right)^{13}=\left(-3\right)^{5^{13}}=\left(-3\right)^{65}\)
\(\Rightarrow\left(-27\right)^{27}< \left(-243^{13}\right)\)
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 \(25^{100}< 27^{100}\)
\(\Rightarrow5^{199}< 3^{300}\)
\(\Rightarrow\dfrac{1}{5^{199}}>\dfrac{1}{3^{300}}\)
\(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\frac{1^7}{81^7}=\frac{1}{\left(3^4\right)^7}=\frac{1}{3^{28}}>\frac{1}{3^{30}}=\frac{1}{\left(3^5\right)^6}=\frac{1^6}{243^6}=\left(\frac{1}{243}\right)^6\)
=>\(\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)
Ta có :
269 < 279 = ( 33 )9 = 327
837 > 817 = ( 34 )7 = 328
MÀ 269 < 327 < 328 < 837 = > 269 < 837
Vậy 269 < 837
ta có : \(\left(-\frac{1}{2}\right)^{500}=\left[\left(-\frac{1}{2}\right)^5\right]^{100}=\left(-\frac{1}{32}\right)^{100}\)
=> \(\left(-\frac{1}{16}\right)^{100}< \left(-\frac{1}{32}\right)^{100}\)
<=> \(\left(-\frac{1}{16}\right)^{100}< \left(-\frac{1}{2}\right)^{500}\)
câu b cũng tương tự nha tất cả đưa về cơ số là -2
\(\left(\frac{1}{243}\right)^9=\frac{1}{243^9}=\frac{1}{\left(3^5\right)^9}=\frac{1}{3^{45}}\)
\(\left(\frac{1}{83}\right)^{13}< \left(\frac{1}{81}\right)^{13}=\frac{1}{81^{13}}=\frac{1}{\left(3^4\right)^{13}}=\frac{1}{3^{52}}\)
Có \(3^{45}< 3^{52}\Rightarrow\frac{1}{3^{45}}>\frac{1}{3^{52}}\)
suy ra \(\left(\frac{1}{243}\right)^9>\left(\frac{1}{83}\right)^{13}\).