a , tổng các phân số đã cho là : 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 = 79/64 

b, \(\frac{79}{64}\)và \(\frac{2017}{2018}\)=  \(\frac{159422}{129152}\)và \(\frac{129088}{129152}\)= \(\frac{159422}{129152}\)> \(\frac{129088}{129152}\)

=> \(\frac{79}{64}\)> \(\frac{2017}{2018}\) 

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

=32/64 + 16/64 + 8/64 + 4/64 + 2/64

=32+16+8+4+2/64 = 66/64= 33/32

b) ta có 33/32 > 1 và 2017/2018<1

nên 33/32 > 2017/2018

Bạn khanh cuong ơi bài này tính đâu đơn giản thế đâu. Cậu làm vắn tắt vậy ai mà hiểu?

Cộng thêm 1/2 vào biểu thức đã cho, có:

S + 1/2= 1/2+1/4+ 1/8+ 1/16+1/32+1/64+1/128

Nhận xét:

a ) \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\times1280\)

= \(\frac{1}{2}\times1280+\frac{1}{4}\times1280+\frac{1}{8}\times1280+\frac{1}{16}\times1280+\frac{1}{32}\times1280+\frac{1}{64}\times1280\)\(+\frac{1}{128}\times1280\)

= 640 + 320 + 160 + 80 + 40 + 20 + 10

= ( 640 + 160 ) + ( 320 + 80 ) + ( 40 + 20 + 10 )

= 800 + 400 + 70

= 1270

b ) ( 2019 - 2017 ) + ( 2015 - 2013 ) + ... + ( 7 - 5 ) + ( 3 - 1 )  (có 505 cặp số )

= 2 + 2 + ... + 2 + 2 ( có 505 số 2 )

= 2 x 505

= 1010

Ta có : A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64

=> 2A  = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32

=> 2A - A = 1 - 1/64

=> A = 1 - 1/64

=> A = 63/64

Ta có : \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{64}\)

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^6}\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^5}\)

\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^5}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^6}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2^6}=1-\dfrac{1}{64}=\dfrac{63}{64}\)

\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...+\dfrac{1}{32}-\dfrac{1}{64}\)

\(=1-\dfrac{1}{64}\)

\(=\dfrac{63}{64}\)

63/64 like nhé

nhân A lên 2A sau đó lấy 2A-A là đc :

2A =1+1/2+.....+1/32

2A-A=(1+1/2+.....+1/32)-(1/2+1/4+.....+1/32+1/64)

A=1-1/64

A=63/64

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(2A=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\times2\)

\(2A=\frac{1}{2}\times2+\frac{1}{4}\times2+\frac{1}{8}\times2+\frac{1}{16}\times2+\frac{1}{32}\times2+\frac{1}{64}\times2\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)

\(A=1-\frac{1}{64}\)

\(A=\frac{63}{64}\)

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(\Rightarrow2A-A=1-\frac{1}{128}=\frac{127}{128}\)

=127/128 nhe