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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: \(2022\cdot2023^{2024}+2023^{2024}=2023^{2024}\left(2022+1\right)\)
\(=2023^{2025}\)
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
Ta có: \(x+2x+3x+4x+...+100x+50=5200\)
\(\Leftrightarrow x\left(1+2+3+4+...+100\right)=5150\)
\(\Leftrightarrow x\cdot\frac{\left(1+100\right)\cdot\left[\left(100-1\right)\div1+1\right]}{2}=5150\)
\(\Leftrightarrow x\cdot5050=5150\)
\(\Rightarrow x=\frac{103}{101}\)
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=\(103\over 101\)
\(x+2x+3x+4x+...+100x+50=5200\)
\(x\left(1+2+3+...+100\right)+50=5200\)
\(5050x+50=5200\)
\(5100x=5200\)
\(x=\frac{52}{51}\)
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