A = \(\frac{24}{48}\)+ \(\frac{12}{48}\)+ \(\frac{8}{48}\)+ \(\frac{2}{48}\)+ \(\frac{1}{48}\)

A = \(\frac{24+12+8+2+1}{48}\)= \(\frac{47}{48}\)

ai tốt bụng thì tk cho mk nha

tui nhìn ko có quy luật j cả

chỉ còn cách là quy đồng (ko ra quy luật)

:):):):):):):):):):):):):):):):) mình đâu có biết!

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

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

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

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

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

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

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

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

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

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

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

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

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

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=1-\frac{1}{100}\)

\(B=\frac{99}{100}\)

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\) 

  \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

  \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

  \(=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+...+\left(-\frac{1}{99}+\frac{1}{99}\right)-\frac{1}{100}\)

    \(=1-\frac{1}{100}=\frac{99}{100}\)

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)

\(\Rightarrow B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow B=1-\frac{1}{100}\)

\(\Rightarrow B=\frac{100}{100}-\frac{1}{100}\)

\(\Rightarrow B=\frac{99}{100}\)

Vậy \(B=\frac{99}{100}\)

tầm là sai ế mình làm bài này nhưng ko ra kq 

TẦM NHƯ HƠI CĂNG

\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+....+\frac{1}{1999}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{1+\left(\frac{1998}{2}+1\right)+\left(\frac{1997}{3}+1\right)+....+\left(\frac{1}{1999}+1\right)}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{2000}{2}+\frac{2000}{3}+\frac{2000}{4}+....+\frac{2000}{2000}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{2000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}\)

\(=\frac{1}{2000}\)

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