so sánh 2^52 và 3^33
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\(a,\) Ta có : \(\hept{\begin{cases}2^{10}=2^{10}\\3^{12}=3^{10}.3^2\end{cases}}\)
Vì \(3^{10}>2^{10}\Rightarrow2^{10}< 3^{10}.3^2\)
Hay \(2^{10}< 3^{12}\)
\(b,\) Ta có : \(\hept{\begin{cases}33^{52}=\left(33^4\right)^{13}=1185921^{13}\\44^{39}=\left(44^3\right)^{13}=85184^{13}\end{cases}}\)
Vì \(1185921^{13}>85184^{13}\)
Do đó : \(33^{52}>44^{39}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(33^{52}=\left(33^4\right)^{13}=\left[\left(3.11\right)^4\right]^{13}=\left(3^4.11^4\right)^{13}=\left(11^3.891\right)^{13}\)
\(44^{39}=\left(44^3\right)^{13}=\left[\left(11.4\right)^3\right]^{13}=\left(11^3.4^3\right)^{13}=\left(11^3.64\right)^{13}\)
Do 891 > 64 => 33^52 > 44^39
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3^{-200}=\left(3^{-2}\right)^{100}=\left(\frac{1}{9}\right)^{100}\)
\(2^{-300}=\left(2^{-3}\right)^{100}=\left(\frac{1}{8}\right)^{100}\)
\(\frac{1}{9}< \frac{1}{8}\Rightarrow\left(\frac{1}{9}\right)^{100}< \left(\frac{1}{8}\right)^{100}\Rightarrow3^{-200}< 2^{-300}\)
\(33^{52}=\left(33^4\right)^{13}\)
\(44^{39}=\left(44^3\right)^{13}\)
\(33^4=\left(33^{\frac{4}{3}}\right)^3\approx106^3\)
\(106^3>44^3\Rightarrow\left(33^4\right)^{13}> \left(44^3\right)^{13}\Rightarrow33^{52}>44^{39}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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3^-200=3^(-2x100)
2^-300=2^(-3x100)
=2^-300>3^-200
chúc bn học tốt
a, 3^(−200) và 2^(−300)
Ta có :
3^(−200) =(3^−2)^100=(1/9)^100
2^(−300) =(2^−3)^100=(1/8)^100
Do 1/9<1/8 nên 3^(−200) < 2^(−300)
b, 33^52 và 44^39
Ta có :
33^52 = ( 33^4)^13
44^39 = ( 44^3 )^13
33^4 = ( 33 4/3 )^3 = 106^3
106^3 > 44^3 ⇒ ( 33^4)^13 > ( 44^3 )^13 ⇒ 33^52 >44^39
#Học tốt#
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22^3^2 : 33^2^3 = 22/33
22/33 = 2/3
vậy 22^3^2 = 2/3 x 33^2^3
suy ra 22^3^2 < 33^2^3