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b.Ta có : = (111.3)111.4 = ( 1114 . 34 )111=
= ( 111 . 4 )111.3 = ( 1113.43)111 =
Vì (1114.81)111 > ( 1113.64 )111 => 333444 > 444333
a. 1030 = ( 103 )10=100010
2100 = ( 210 )10=102410
Vì 100010<102410 nên 1030<2100

Nguyễn Khánh Phương
Bài 1 :
a) 149 - ( 35 : x + 3 ) x 17 = 13
( 35 : x + 3 ) x 17 = 149 - 13
( 35 : x + 3 ) x 17 = 136
( 35 : x + 3 ) = 136 : 17
( 35 : x + 3 ) = 8
35 - x = 8 - 3
35 - x = 5
x = 35 - 5
x = 30
b, 121 : 11 − ( 4x + 5 ) : 3 = 4
11 − 4x + 5 : 3 = 4
4x + 5 : 3 = 11 − 4
4x + 5 : 3 = 7
4x + 5 = 7 x 3
4x + 5 = 21
4x = 21 − 5
4x = 16
x = 16 : 4
x = 4

a)\(3^2.9^3=9.9^3=9^{1+3}=9^4\)
b)\(2^2.5^2=4.25=100=10^2\)
c)\(8^5.2^3=8^5.8=8^{5+1}=8^6\)
d)\(9^8:3^2=9^8:9=9^{8-1}=9^7\)

a,320 và 274
320=(35)4=2434>274
Vậy 320>274
b,534 và 25x530
25x530=52x530=532<534
=>534>25x530.
c,224và 266
224=(24)6=166<266
=>224<266
d,1030và 450
1030=(103)10=100010
450=(45)10=102410
Vì 100010<102410nên 1030<450.
e,2300và 3200
2300=(23)100=8100
3200=(32)100=9100
Vì 8100<9100 nên 2300<3200

2 mũ 5 bằng 32
3 mũ 3 bằng 27
5 mũ 2 bằng 25
10 mũ 9 bằng 1000000000

Bài 1: a) \(M=1+5+5^2+...+5^{100}\)
\(5M=5+5^2+5^3+...+5^{101}\)
\(5M-M=\left(5+5^2+5^3+...+5^{101}\right)-\left(1+5+5^2+...+5^{100}\right)\)
\(4M=5^{101}-1\)
\(M=\frac{5^{101}-1}{4}\)
b) \(N=2+2^2+...+2^{100}\)
\(2N=2^2+2^3+...+2^{101}\)
\(2N-N=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)
\(N=2^{101}-2\)
Bài 2:
a) \(16^{32}=\left(2^4\right)^{32}=2^{128}\)
\(32^{16}=\left(2^5\right)^{16}=2^{80}\)
Vì \(2^{128}>2^{80}\Rightarrow16^{32}>32^{16}\)

a,\(2^4\cdot3^5:6^4\)
\(=\frac{2^4\cdot3^6}{\left(2\cdot3\right)^4}\)
\(=\frac{2^4\cdot3^6}{2^4\cdot3^4}\)
\(=3^2\)
Bài 2
\(a,5^3\cdot8=5^3\cdot2^3=10^3=1000\)
\(b,2^5-2019^0=32-1=31\)
\(c,3^3+2^5-1^{10}=27+32-1=58\).
\(d,9^2\cdot33-81\cdot23+5^2=81\cdot33-81\cdot23+25\)
\(=81\cdot\left(33-23\right)+25\)
\(=810+25=835\)
\(g,\left[2^2+6^2\right]:5+11^2\)
\(=\left[4+36\right]:5+121\)
\(=40:5+121=8+121\)
\(=129\)
\(d,\frac{14\cdot3^{10}-5\cdot3^{10}}{3^{12}}\)
\(=\frac{3^{10}\cdot\left(14-5\right)}{3^{12}}\)
\(=\frac{3^{10}\cdot9}{3^{12}}\)
\(=\frac{3^{10}\cdot3^2}{3^{12}}=\frac{3^{12}}{3^{12}}\)
\(=1\)

\(23+3x=5^6:5^3\) \(9^{x-1}=9\) \(x^4=16\) \(2^x:2^5=1\)
\(23+3x=5^2\) => x-1= 0 \(x^4=2^4\) \(2^x=1.2^5\)
\(23+3x=25\) x= 0+1 => x= 2 \(2^x=2^5\)
3x= 25-23 x= 1 => x= 5
3x= 2
x= \(\frac{2}{3}\)
Đặt A = 52 + 55 + ... + 5100
=> 53A = 55 + 58 + ... + 5103
=> 125A - A = ( 55 + 58 + ... + 5103 ) - ( 52 + 55 + ... + 5100 )
=> 124A = 5103 - 52
=> A = 5103 - 52 / 124
Đặt \(B=5^2+5^5+...+5^{100}\)
\(\Rightarrow5^3B=5^5+5^8+...+5^{103}\)
\(\Rightarrow125B-B=\left(5^5+5^8+...+5^{103}\right)-\left(5^2+5^5+...+5^{100}\right)\)
\(\Rightarrow124B=5^{103}-5^2\)
\(\Rightarrow B=5^{103}-\frac{5^2}{124}\)