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Ta có: `8^111 =(2^3 )^111 =2^(3.111)=2^333`
`4^170 =(2^2 )^170 =2^(2.170)=2^340`
Vì `333<340=>8^111 <4^170`
Ta có: `3^300 =3^(3.100)=(3^3 )^100=27^100`
`5^200 =5^(2.100)=(5^2 )^100 =25^100`
Vì `27>25=>3^300 >5^200`
a: 8^111=2^333
4^170=(2^2)^170=2^340
mà 333<340
nên 8^111<4^170
b: 3^300=(3^3)^100=27^100
(5^200)=(5^2)^100=25^100
mà 27>25
nên 3^300>5^200
![](https://rs.olm.vn/images/avt/0.png?1311)
a>ƯCLN
1230=2.3.5.41
4800=26.3.52
ƯC{1230;4800}=2.3=6={1;2;3;6}
=>ƯCLN{1230;4800}=6
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Ta có: \(x+2x+3x+4x+...+100x+50=5200\)
\(\Leftrightarrow x\left(1+2+3+4+...+100\right)=5150\)
\(\Leftrightarrow x\cdot\frac{\left(1+100\right)\cdot\left[\left(100-1\right)\div1+1\right]}{2}=5150\)
\(\Leftrightarrow x\cdot5050=5150\)
\(\Rightarrow x=\frac{103}{101}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(12+2^2+3^2+4^2+5^2\)
\(=12+4+9+16+25\)
\(=16+50=66\)
\(\left(1+2+3+4+5\right)^2=15^2=225\)
=>\(12+2^2+3^2+4^2+5^2< \left(1+2+3+4+5\right)^2\)
b: \(1^3+2^3+3^3+4^3=\left(1+2+3+4\right)^2< \left(1+2+3+4\right)^3\)
c: \(5^{202}=5^2\cdot5^{200}=25\cdot5^{200}>16\cdot5^{200}\)
d: \(18\cdot4^{500}=18\cdot2^{1000}\)
\(2^{1004}=2^4\cdot2^{1000}=16\cdot2^{1000}\)
=>\(18\cdot4^{500}>2^{1004}\)
e: \(2022\cdot2023^{2024}+2023^{2024}=2023^{2024}\left(2022+1\right)\)
\(=2023^{2025}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=3+3^2+3^3+...+3^{300}\)
\(B=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{298}+3^{299}+3^{300}\right)\)
\(B=\left(3+3^2+3^3\right)+3^3\cdot\left(3+3^2+3^3\right)+...+3^{297}\cdot\left(3+3^2+3^3\right)\)
\(B=39+3^3\cdot39+...+3^{297}\cdot39\)
\(B=39\cdot\left(1+3^3+...+3^{297}\right)\)
Vậy B chia hết cho 39
![](https://rs.olm.vn/images/avt/0.png?1311)
(3+X)+(6+X)+(9+X)+…+(99+X)=3300
(3+6+9+…+99)+(X+X+X+…+X)=3300
A
Số số hạng dãy A là: (99-3):3+1=33
A=(99+3)x33:2=(99+3):2x33=51x33
51x33+33xX=3300
33x(51+X)=3300
51+X=3300:33
51+X=100
X=100-51
X=49
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x+x.2+x.3+...+x.100+50=5200
x.(1+2+3+..+100)+50=5200
Có 1+2+3+...+100=(100+1).100:2=5050
suy ra x.(1+2+3+...+100)+50=x.5050+50=5200
x=5150:5050=\(103\over 101\)
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a: 43/52>26/52=1/2=60/120
b: 17/68=1/4<1/3=35/105<35/103
c: \(\dfrac{2018\cdot2019-1}{2018\cdot2019}=1-\dfrac{1}{2018\cdot2019}\)
\(\dfrac{2019\cdot2020-1}{2019\cdot2020}=1-\dfrac{1}{2019\cdot2020}\)
2018*2019<2019*2020
=>-1/2018*2019<-1/2019*2020
=>\(\dfrac{2018\cdot2019-1}{2018\cdot2019}< \dfrac{2019\cdot2020-1}{2019\cdot2020}\)
5200 = (52)100 = 25100
3300 = (33)100 = 27100
vì: 27100 > 25100
=> 5200 < 3300