dễ quá hen

= 4/5 + 1/5 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2

= (4/5 + 1/5) + (1/2 + 1/2) + (1/2 + 1/2) + 1/2
= 1 + 1 + 1 + 1/2
= 3 + 1/2
= 6/2 + 1/2 (Hoặc ra luôn là hỗn số 3 và 1/2)
= 7/2

bằng 15/16 nhé. 

Không nhé !

\(\frac{13}{32}\)+ \(\frac{34}{64}\)= \(\frac{15}{16}\)

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

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

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

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

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

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

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

\(=\frac{64}{64}-\frac{1}{64}=\frac{63}{64}\)

63 phan 96

Phân số đó là \(\frac{63}{64}\)

=127/128

~ chúc bn hok tốt ~

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

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

\(=\frac{126}{128}=\frac{63}{64}\)

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

\(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)

\(=\frac{1+1+1+1+1+1+1}{2}\)

\(=\frac{7}{2}\)

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

\(T=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{64}-\frac{1}{128}\right)\)

\(\Rightarrow T=1-\frac{1}{128}=\frac{127}{128}\)

ta có:A=1/2+1/2^2+1/2^3+...+1/2^6+1/2^7  (1)

 2A=     2.(1/2+1/2^2+1/2^3+...+1/2^6+1/2^7)

      =1+1/2+1/2^2+....+1/2^6+1/2^7  (2)

  lấy (2) trừ (1) vế với vế ta được:
2A-A=(1+1/2+1/2^2+....+1/2^6+1/2^7)-(1/2+1/2^2+...+1/2^6+1/2^7)

     A=1-1/2^7

  VẬY A=1-1/2^7

\(\frac{127}{128}\)

\(A\cdot2=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{256}\right)\cdot2\)

\(=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}\)

\(A\cdot2-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)

\(A=1-\frac{1}{256}=\frac{255}{256}\)

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

\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)

\(2A=1+\frac{1}{2}+...+\frac{1}{2^7}\)

\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)

\(A=1-\frac{1}{2^8}\)

\(A=\frac{2^8-1}{2^8}\)

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

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

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

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

\(A=1-\frac{1}{64}=\frac{63}{64}\)

Ta thấy: 1/2 = 1 - 1/2

1/2 + 1/4 = 3/4 = 1- 1/4

1/2 + 1/4 + 1/8 = 7/8 = 1 - 1/8

Tương tự ta có:

1/2 + 1/4 +1/8 + 1/16 + 1/32 + 1/64 = 1 - 1/64 = 63/64

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