Ta có 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64

= ( 1/2 - 1/4 ) + ( 1/8 - 1/16 ) + ( 1/32 -  1/64)

= 1/4 + 1/16 + 1/64 

= 16 + 4 + 1 /64

= 21/64 < 21/63 = 1/3 

Vậy 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 < 1/3 ( đpcm ) Chúc bn hok tốt  . k mik nha 

Ta có :

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

\(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}< \frac{1}{3}\)

\(\frac{16}{64}+\frac{4}{64}+\frac{1}{64}< \frac{1}{3}\)

\(\frac{16+4+1}{64}< \frac{1}{3}\)

\(\frac{21}{64}< \frac{1}{3}\)

=> 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 < 1/3

Ta có: \(\frac{1}{2}-\frac{1}{4}=\frac{2}{4}-\frac{1}{4}=\frac{2-1}{4}=\frac{1}{4}\)

\(\frac{1}{8}-\frac{1}{16}=\frac{2}{16}-\frac{1}{16}=\frac{2-1}{16}=\frac{1}{16}\)

\(\frac{1}{32}-\frac{1}{64}=\frac{2}{64}-\frac{1}{64}=\frac{2-1}{64}=\frac{1}{64}\)

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

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

=\(\frac{16}{64}+\frac{4}{64}+\frac{1}{64}=\frac{21}{64}\)

Ta có: \(\frac{21}{64}< \frac{21}{63}=\frac{1}{3}\)

=> \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)

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

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

\(A+2A=1-\frac{1}{64}\)

\(3A=1-\frac{1}{64}< 1\)

=>A<1/3

=>đpcm

1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 = 21/64

Nên 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 < 1/3

Chúc bạn học giỏi

\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-...-\frac{1}{64}=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-..-\frac{1}{2^6}\) = A 

2A = 1 - 1/2 + 1/2^2 - ... - 1/2^5

2A + A = 1 - 1/2 + 1/2^2 - ... - 1/2^5 + 1/2 - 1/2^2 + 1/2^3 - 1/2^4 - .. - 1/2^6

     3A  = \(1-\frac{1}{2^6}=\frac{2^6-1}{2^6}\)

hơi đúng

Đặt A = 1/2 − 1/4 + 1/8 − 1/16 + 1/32 − 1/64 A = 1/2 − 1/4 + 1/8 − 1/16 + 1/32 − 1/64
2A = 1 − 1/2 + 1/4 − 1/8 + 1/16 − 1/322 A =1 − 1/2 + 1/4 − 1/8 + 1/16 − 1/32
3A = 2A + A = 1 − 1/64 < 1 ⇒ A < 1/3 .

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

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

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

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

\(=\frac{21}{64}< \frac{1}{3}\)(đpcm)