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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#
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)2^{300}=(2^3)^100=8^100`
`3^200=(3^2)^100=9^100`
Vì `9^100>8^100`
`=>2^300<3^200`
`b)3xx24^10`
`=3.(3.8)^10`
`=3^{11}.8^10`
`=3^{11}.2^30`
`2^300=2^{30}.2^{270}`
`=2^{30}.8^{90}`
Vì `3^11<8^90`
`=>3^{11}.2^30<8^{90}.2^30=2^300`
`=>3xx24^{10}<2^300+3^20+4^30`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,2^{24}=\left(2^3\right)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8>8^8\)
\(\Rightarrow3^{16}>2^{24}\)
\(b,2^{300}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=\left(3^2\right)^{100}=9^{100}>8^{100}\)
\(\Rightarrow3^{200}>2^{300}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b, 2300=23.100=[23]100=8100
3200=32.100=[32]100=9100
=> 8100 < 9100 . Vậy 2300 < 3100
![](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)
2300 và 3200
2300 = ( 23 )100 = 8100
3200 = ( 32 )100 = 9100
Vì 8 < 9 nên 2300 < 3200
2300 = ( 23)100 = 8100
3200 = ( 32)100 = 9100
Vì 8100 < 9100 nên 2300 < 3200
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: Sửa đề: 1/3^200
1/2^300=(1/8)^100
1/3^200=(1/9)^100
mà 1/8>1/9
nên 1/2^300>1/3^200
b: 1/5^199>1/5^200=1/25^100
1/3^300=1/27^100
mà 25^100<27^100
nên 1/5^199>1/3^300
Ta có:
2300=(23)100=8100
3200=(32)100=9100
Vì 8<9 nên 8100<9100
Vậy 2300<3200
Suy ra A<B
Ta có:
\(A=2^{300}\)\(=\left(2^3\right)^{100}\)\(=8^{100}\)
\(B=3^{200}\)\(=\)\(\left(3^2\right)^{100}\)\(=9^{100}\)
Vì \(8^{100}< 9^{100}\)nên \(A< B\)