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\(A=10^{25}+\frac{1}{10^{26}}+1=1\cdot10^{25}\)
\(B=10^{26}+\frac{1}{10^{27}}+1=1\cdot10^{26}\)
\(1\cdot10^{25}< 1\cdot10^{26}\Rightarrow A< B\)
a) 27/82 < 26/75 ( 2025/6250 < 2132\6250)
b) -49/78 > 64/ -95 ( - 3136/7410 > -4992/7410)
c) ta có: \(A=\frac{54.107-53}{53.107}=\frac{53.107+(107-53)}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}=\frac{134.269+\left(269-133\right)}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)
\(\Rightarrow A< B\)
d) ta có: \(A=\frac{3^{10}+1}{3^9+1}=\frac{3.\left(3^9+1\right)-2}{3^9+1}=\frac{3.\left(3^9+1\right)}{3^9+1}-\frac{2}{3^9+1}=3-\frac{2}{3^9+1}\)
\(B=\frac{3^9+1}{3^8+1}=\frac{3.\left(3^8+1\right)-2}{3^8+1}=\frac{3.\left(3^8+1\right)}{3^8+1}-\frac{2}{3^8+1}=3-\frac{2}{3^8+1}\)
mà \(\frac{2}{3^9+1}< \frac{2}{3^8+1}\Rightarrow3-\frac{2}{3^9+1}< 3-\frac{2}{3^8+1}\)
=> A < B
a) ta thấy
2638=(2.13)38=238.1338
6425=(23.8)25=228.825
ta dễ dàng nhận thấy: 228<238 và 825<1338 nên:
2638>6425
a)1733>3118
b)24317>8222
c)20160<39945
d)2233>3322
e)222333<333222
F)<
G)>
BÀI 1: giải :
A=25.33-10
=25.(31+2)-10
=25.31+25.2-10
=25.31+(25.2-10)
=25.31+40
B=31.26+10
=31(25+1)+10
=31.25+35+10
=31.25+(35+10)
=31.25+45
Vì 45>40 nên B>A
vậy :A=25.33-10 <B=31.26+10
bài 2:
Đặt A= 1+2+22+23+....+299+2100
=>2A=2+2^2+2^3+...+2^100+2^101
=>2A-A=2+2^2+2^3+...+2^100+2^101-(1+2+2^2+2^3+...+2^99+2^100)
=>A=2+2^2+2^3+...+2^100+2^101-1-2-2^2-2^3-...-2^99-2^100=2^101-1
=>A= 2^101 -1
A=25.33-10
=25.(31+2)-10
=25.31+25.2-10
=25.31+(25.2-10)
=25.31+40
B=31.26+10
=31(25+1)+10
=31.25+35+10
=31.25+(35+10)
=31.25+45
Vì 45>40 nên B>A
vậy :A=25.33-10 <B=31.26+10
a)Ta có: \(2^{91}=2^{90}x2=\left(2^6\right)^{15}x2=64^{15}x2\)
Vì 6415 >515 =>6415 x 2 >515
Hay 2 91 > 515
b)Ta có: 521=7812514
vì 7812514>2614nên 521>2614
Ta có:
2623 < 2723 => 2623 < (33)33 => 2623 < 399
8225 > 8125 => 8225 > (34)25 => 8225 > 3100
Dễ dàng ta thấy : 2623 < 8225
Vậy A < B