\(C=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)

\(2C=\frac{4}{3}+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}\)

\(2C-C=\frac{4}{3}-\frac{2}{192}\)

\(C=\frac{127}{96}\)

\(C=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)

\(C=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)

\(C=\frac{254}{192}=\frac{127}{96}\)

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)

=\(\frac{2}{3}+\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)

    Ta xét phần trong ngoặc :

\(\frac{1}{2}+\frac{1}{4}=1-\frac{1}{4}\)

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

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

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

=> \(1+\frac{63}{64}=\frac{127}{96}\)

Vậy tổng trên có kết quả là \(\frac{127}{96}\)

k nha

Ta thấy tất cả các phân số đều có mẫu chung là 192

=> \(\frac{128+64+32+16+8+4+2}{192}\)

= \(\frac{254}{192}\)= \(\frac{127}{96}\)

2/3+2/6+2/12+2/24+2/48+2/96+2/192

=2/3+1/3+1/6+1/12+1/24+1/48+1/96

=127/96

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{29}{96}+\frac{2}{192}\)

\(=\frac{2}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)

\(=\frac{254}{192}=\frac{127}{96}\)

Đặt bt bằng A

Ta có 2A= 2(2/3 + 2/6 + 2/12 +2/24 + 2/48 + 2/96 + 2/192)

2A= 4/3 +2/3 + 2/6 + 2/12 + 2/24 + 2/48 + 2/96

A= 2A-A= (4/3 +2/3 + 2/6 + 2/12 + 2/24 + 2/48 + 2/96) - (2/3 + 2/6 + 2/12 +2/24 + 2/48 + 2/96  2/192)

A=4/3 +2/3 + 2/6 + 2/12 + 2/24 + 2/48 + 2/96 - 2/3 - 2/6 - 2/12 - 2/24 - 2/48 - 2/96 - 2/192

A=(2/3 - 2/3) + (2/6 - 2/6) + ( 2/12 -  2/12) + (2/24 - 2/24) + (2/48 - 2/48) + ( 2/96 -  2/96) + (4/3 - 2/192)

A=0+0+0+0+0+0+ (256/192 - 2/192)

A=254/192

A=127/96(rút gọn phân số) 

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{192}\)

\(=2\times\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{192}\right)\)

\(=2\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{96}-\frac{1}{192}\right)\)

\(=2\times\left(1-\frac{1}{192}\right)\)

\(=2\times\frac{191}{192}\)

\(=\frac{382}{192}=\frac{191}{96}\)

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)

\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+...+\frac{1}{96}-\frac{1}{192}\right)\)

\(=2.\left(1-\frac{1}{192}\right)\)

\(=2.\frac{191}{192}\)

\(=\frac{382}{192}=\frac{191}{96}\)

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+...+\frac{2}{192}.\)

\(=2\times\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{192}\right)\)

\(=2\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{96}-\frac{1}{192}\right)\)

\(=2\times\left(1-\frac{1}{192}\right)\)

\(=2\times\frac{191}{192}=\frac{191}{68}\)

\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+...+\frac{2}{192}\)

\(=\frac{1}{3.1}+\frac{1}{3.2}+\frac{1}{3.2^2}+...+\frac{1}{3.2^6}\)

\(=\frac{1}{3}.\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)

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

\(\Rightarrow2A=2.\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)

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

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

\(\Rightarrow A=2-\frac{1}{2^6}=2-\frac{1}{64}=\frac{127}{64}\)

\(\Rightarrow\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+...+\frac{2}{192}=\frac{1}{3}.\frac{127}{64}=\frac{127}{192}\)

két quả là:125/96

2/3+ 2/6+ 2/12+ 2/24+ 2/48+ 2/192= 2/3+ 1/3+ 1/6+1/12+1/24+1/96

                                                        =64/96+32/96+16/96+8/96+4/96+1/96

                                                        =125/96

\(=\frac{125}{96}\)

Câu 1

Ta có \(\frac{119x83-183}{120x83-266}=\frac{119x83-183}{119x83+83-266}=\frac{119x83-183}{119x83-183}=1\) 

cái a bằng 1962

cái b bằng 127/192

à quên mình chưa rút gọn phân số đấy đâu bạn ạ

ban rút gọn phân số đấy hộ mình nha

bạn giải từng bước ra giúp mình nhé

   \(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}\)+\(\frac{2}{96}\)

=\(2\)x (\(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}\)\(+\frac{1}{96}\))

=\(2\)x (\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+...\)\(+\frac{1}{48}-\frac{1}{96}\))

=\(2\)x (\(1-\frac{1}{96}\))

=\(2\)x \(\frac{95}{96}\)

=\(\frac{190}{96}=\frac{95}{48}\)

Sao mà mình hỏi bài này từ lâu lắm rồi mà vẫn chưa có bạn nào trả lời nhỉ?

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

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

2A-A = \(1-\dfrac{1}{32}\)

A=  \(\dfrac{31}{32}\)

b)\(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\) 

Dặt A=

3A= \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

3A-A=\(1-\dfrac{1}{792}\) 

2A= \(\dfrac{791}{792}\) 

A= \(\dfrac{791}{792}:2=\dfrac{791}{1584}\)