\(a,2^x.4=128\)

\(\Leftrightarrow2^x=128:4\)

\(\Leftrightarrow2^x=32\)

\(\Leftrightarrow2^x=2^5\)

\(\Rightarrow x=5\)

\(b,2^x=4.128\)

\(\Leftrightarrow2^x=2^2.2^7\)

\(\Leftrightarrow2^x=2^9\)

\(\Rightarrow x=9\)

\(c,3^x=3^3.3^5\)

\(\Leftrightarrow3^x=3^8\)

\(\Rightarrow x=8\)

\(d,2^x.\left(2^2\right)^3=\left(2^3\right)^2\)

\(\Leftrightarrow2^x.2^6=2^6\)

\(\Leftrightarrow2^x=2^6:2^6\)

\(\Leftrightarrow2^x=1\)

\(\Rightarrow x=0\)

Trả lời:

a, 2^x . 4 = 128

=> 2^x = 32

=> 2^x = 2⁵

=> x = 5

b, 2^x = 4.128

=> 2^x = 512

=> 2^x = 2⁹

=> x = 9

c, 3^x = 3³ . 3⁵ 

=> 3^x = 3⁸

=> x = 8

d, 2^x ( 2² )³ = ( 2³ )²

=> 2^x . 2⁶ = 2⁶

=> 2^x = 1

=> 2^x = 2⁰

=> x = 0

a=128:4x3=96

Ta có :

\(A=\frac{3}{2}+\frac{3}{4}+...+\frac{3}{128}\)

\(2A=3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\frac{3}{32}+\frac{3}{64}\)

\(2A-A=\left(3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\frac{3}{32}+\frac{3}{64}\right)-\left(\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\frac{3}{32}+\frac{3}{64}+\frac{3}{128}\right)\)

\(A=3-\frac{3}{128}\)

\(A=\frac{381}{128}\)

Ủng hộ mk nha !!! ^_^

A=3:(2+4+8+16+32+64+128)

A=3:254

A=\(\frac{3}{254}\)

tại sao z ???????

a) 128 - 3 ( x + 4 ) = 23

     3 ( x + 4 ) = 128 - 23

     3 ( x + 4 ) = 105

     x + 4 = 105 : 3

     x + 4 = 35

     x = 35 - 4

     x = 31

Học tốt bạn nhé!!!

Arigatou ^^

b) \(\left(12x-64\right).8^3=4.8^4\)

\(\left(12x-64\right).8^3=32.8^3\)

\(\Rightarrow12x-64=32\)

12x = 32 + 64

12x  = 96

x = 96 : 12

x = 8

2A - A= 2²¹chia hết cho 2⁷

suy ra A chia hết cho 128

2A-A=2²¹ chia het cho2⁷,suy ra Achia het cho 128

2A-A=2\(^{21}\)chia hết cho2\(^7\)

=>A chia hết cho 128

a) 128 - 3( x + 4 ) = 23

<=> 3x + 12 = 128 - 23 =105

<=> 3x = 105 - 12 = 93

<=> x = 31

b) [ ( 6x - 39 ) : 7 ] . 4 =12

<=> ( 6x - 39 ) : 7 = 12 / 4 =3

<=> 6x - 39 = 3 . 7 =21

<=> 6x = 21 + 39 = 60

<=> x =10 

A=4+2^2+2^3+.......+2^20 .Hỏi A có chia hết cho 128 không?
A = 4 + 4 + 2^3+.......+2^20
A = 2^3 + 2^3+ 2^4 + 2^5 +.......+2^20
A = 2.2^3 + 2^4 + 2^5 +2^6 + 2^7 + .......+2^20
A = 2^4 + 2^4 + 2^5 + 2^6 + 2^7 + .......+2^20
A= 2.2^4 + 2^5 + 2^6 + 2^7 + .......+2^20
A= 2^5 + 2^5 + 2^6 + 2^7 + .......+2^20
A = 2.2^5 + 2^6 + 2^7 + .......+2^20
A = 2^6 + 2^6 + 2^7 + .......+2^20
A= 2.2^6 + 2^7 + 2^8 + .......+2^20
A = 2^7 + 2^7 + 2^8 +.......+2^20
A = 2^7( 1 + 1+ 2 + ....+ 2^13)
A = 128 (1 + 1+ 2 + ....+ 2^13)
Vậy A chia hết cho 128