a, \(\frac{2}{3}+\frac{2}{3}+\frac{6}{3}=\frac{10}{3}\)

b,\(\frac{3}{4}+\frac{3}{4}+\frac{3}{2}=\frac{6}{4}+\frac{3}{2}=\frac{3}{2}+\frac{3}{2}=\frac{6}{2}=3\)

ta có: 1+1/2+2+1/4+...+9+1/512

        =(1+2+3+4+...+9)+(1/2+1/4+...+1/512)

       =45+(1/2+1/4+...+1/512)

gọi số hạng (1/2+1/4+...+1/512) là a ta được :

a=1/2+1/4+...+1/512

2a=1+1/2+1/4+1/8+...+1/256

2a-a=(1+1/2+1/4+...+1/256)-(1/2+1/4+...+1/512)

      =1-1/512

      =511/512

vậy kết quả của biểu thức đó là45+511/512

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

\(\frac{32+16+8+4+2}{64}=\frac{62}{64}=\frac{31}{32}\)

Tk mh nhé , mơn nhìu !!!
~ HOK TỐT ~

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

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

= 63/64

Ta đặt biểu thức trên là A

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

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

A = \(1-\frac{1}{64}\)

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

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}=\frac{481}{280}\)

Bạn phải giải ra 

=127/128

~ chúc bn hok tốt ~

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

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

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

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

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

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

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

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

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

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

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

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 : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)

Đặ A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)(1)

=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)(2)

Lấy (2) trừ (1) theo vế ta có : 

2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)

=> A = \(1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{20}}\)

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)

\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^9}\)

\(\Rightarrow2A-A=1-\frac{1}{2^{10}}=\frac{1023}{1024}\)

Xin lỗi bạn Trần thị mai Chi nha mk bấm sai kết quả . Kết quả đúng là : 

\(A=\frac{2^{10}-1}{2^{10}}\)

tính tử:

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}=\frac{15}{16}\)

Tính mẫu: 

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}=\frac{15}{30}+\frac{5}{30}+\frac{2,5}{30}+\frac{1}{30}=\frac{23,5}{30}=\frac{235}{300}=\frac{47}{60}\)

Có: \(\frac{15}{16}:\frac{47}{60}=\frac{15}{16}.\frac{60}{47}=1\frac{37}{188}\)