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AH
Akai Haruma
Giáo viên
17 tháng 11 2023

Lời giải:

8A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$

$=[(3^2-1)(3^2+1)](3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$

$=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$

$=(3^8-1)(3^8+1)(3^{16}+1)-4.3^{32}$

$=(3^{16}-1)(3^{16}+1)-4.3^{32}$

$=3^{32}-1-4.3^{32}$

$=-3.3^{32}-1=-3^{33}-1$
$\Rightarrow A=\frac{-3^{33}-1}{8}$

b: A=1/3+1/9+...+1/3^10

=>3A=1+1/3+...+1/3^9

=>A*2=1-1/3^10=(3^10-1)/3^10

=>A=(3^10-1)/(2*3^10)

c: C=3/2+3/8+3/32+3/128+3/512

=>4C=6+3/2+...+3/128

=>3C=6-3/512

=>C=1023/512

d: A=1/2+...+1/256

=>2A=1+1/2+...+1/128

=>A=1-1/256=255/256

12 tháng 7 2016

Đặt \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\).Ta có : 

\(=>\left(3-1\right)A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=>2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=>2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

...............................................................................

Cuối cùng \(=>2A=3^{64}-1\).

\(=>A=\frac{3^{64}-1}{2}\)

12 tháng 7 2016

Đặt \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=...........................................\)

\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)

\(\Rightarrow A=\frac{3^{64}-1}{2}\)

20 tháng 3 2016

a)\(\frac{32}{64}-\frac{16}{64}+\frac{8}{64}-\frac{4}{64}+\frac{2}{64}-\frac{1}{64}\le\frac{1}{3}\)

\(\Rightarrow\frac{32-16+8-4+2-1}{64}=\frac{23}{64}\)\

\(\Rightarrow\frac{23}{64}=0,359375;\frac{1}{3}=0,33333...\)

đề sao lạ vậy

20 tháng 3 2016

@ Bùi Long Vũ tinh sai roi kia:

32-16+8-4+2-1=21 mak 

14 tháng 8 2016

1/ \(=-\frac{64}{27}.\frac{243}{32}\)

\(=-\frac{243}{16}\)

2/ \(=\frac{1}{81}.\frac{5361441}{64}\)

\(=\frac{6561}{64}\)

3/ \(=-\frac{2197}{512}.36,71356045\)

\(=-\frac{2048}{13}\)

tíc mình nha

14 tháng 8 2016

1\(\left(-\frac{4}{3}\right)^3.\left(\frac{9}{16}\right)^5=-\frac{2187}{16384}\)

2\(\left(\frac{1}{3}\right)^4.\left(-\frac{9}{2}\right)^6=\frac{6561}{64}\)

3\(\left(-\frac{13}{8}\right)^3.\left(-\frac{32}{13}\right)^4=-41,00457607\)

1 tháng 8 2017

a ) 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256

Đạt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256

A x 2 = 2 x (  1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256)

A x 2 = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128

Lấy A x 2 - A ta có :

A x 2 - A = ​1 + 1/2 + ..... + 1/128 - 1/2 + 1/4 + ........ + 1/256

A x ( 2 - 1 ) = 1 - 1/ 256

A               =           255/256

b) 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

Đặt A = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

A x 3 = 3 x ( 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)

        = 1 + 1/ 3 + 1/9 + 1/27 + 1/81 + 1/243

Lấy A x 3 - A ta có : 

A x 3 - A = 1 + 1/3 + 1/9 +.....  + 1/243 -  1/3 + 1/9 +........+ 1/243 + 1/29

A x ( 3 - 1 ) = 1 - 1/29

A x2          =    28/29

A = 28/29 : 2 ( tự tính

1 tháng 8 2017

Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{256}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{128}\)

\(\Rightarrow2A-A=1-\frac{1}{256}\)

\(\Rightarrow A=\frac{255}{256}\)

23 tháng 7 2018

\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(2B=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(2B=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(2B=\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(B=\dfrac{3^{32}-1}{2}< A=3^{32}-1\)

23 tháng 7 2018

\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =>2B=2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)=3^{32}-1\\ =>A=\dfrac{3^{32}-1}{2}< B\)

5 tháng 7 2017

Nó hơi dài cậu chờ tí nka !

5 tháng 7 2017

Mình ghi nhầm đề bài 1 tí đề bài là :

So sánh 2 số A và B biết : 

A = (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1) và B = 3^32 - 1