MSC:192

\(\frac{4}{3}+\frac{1}{3}+\frac{1}{12}+\frac{1}{48}+\frac{1}{192}\)

\(=\frac{256}{192}+\frac{64}{192}+\frac{16}{192}+\frac{4}{192}+\frac{1}{192}\)

\(=\frac{341}{192}\)

=341/192 k mik  nha***

\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)

=\(\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)

=\(\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{4}-\frac{3}{8}+\frac{3}{8}-\frac{3}{16}+\frac{3}{16}-\frac{3}{36}\)

=\(\frac{3}{1}-\frac{3}{36}\)=\(\frac{35}{12}\)

a)=768/512+192/512+48/512+12/512+3/512

=768+192+48+12+3/512

=1023/512 

b)=405/81+135/81+45/81+15/81+5/81

=405+135+45+15+5/81

=595/81

c)=256/192+64/192+16/192+4/192+1/192

=256+64+16+4+1/192

=341/192

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

\(=\frac{-263}{8085}\)

BN ƠI = \(-\frac{263}{8085}\)NHA Nguyễn Hiền Anh!

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

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

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

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

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

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

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

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

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

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

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

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

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

      \(C=\frac{1990.997}{1994.995}\)

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

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

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

Câu 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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