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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    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

\(C=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(C=\frac{2}{3}+\frac{2}{3.2}+\frac{2}{3.4}+\frac{2}{3.8}+\frac{2}{3.16}+\frac{2}{3.32}+\frac{2}{3.64}\)
\(C=1-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{8}+...+\frac{2}{64}\)
\(C=1-\frac{2}{64}\)
\(C=\frac{31}{32}\)
Làm mò, không biết đúng không nữa?

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

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

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

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

=\(2\)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 \(\frac{95}{96}\)

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

\(B=\)\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)

\(B=\frac{3.1+3.11+3.111+3.1111+3.11111}{4.1+4.11+4.111+4.1111+4.11111}\)

\(B=\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)

\(B=\frac{3}{4}\)

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)

\(A.2=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right).2\)

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

=>\(A.2-A=\left(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right)\)

\(A=\frac{2}{3}-\frac{1}{192}\)

\(A=\frac{127}{192}\)

\(\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)

Đặt \(C=\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)

      \(C=\frac{1995.1990.1997.1993.997}{1997.1993.1994.1995.995}\)

      \(C=\frac{1990.997}{1994.995}\)

      \(C=\frac{995.2+997}{997.2+995}=1\)

\(B=\frac{3+33+333+3333+ 33333}{4+44+444+4444+44444}\)

\(\Rightarrow B=\frac{3\left(1+11+111+1111+11111\right)}{4\left(1+11+111+1111+11111\right)}=\frac{3}{4}\)

B=\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)

B= \(\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)

B=\(\frac{3}{4}\)

Sau mình làm tiếp vội quá! k mình nha