Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)

\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)

\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)

\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

\(A=\frac{B}{6}=\frac{100}{2}=50\)

Vậy \(A=50\)

6 ở đâu hả https://olm.vn/thanhvien/aihaibara0

1)\(\frac{-8}{5}+\frac{207207}{201201}\)

=\(\frac{-8}{5}+\frac{207}{201}\)

=\(\frac{-8}{5}+\frac{69}{67}\)

=\(\frac{-191}{335}\)

A= 1/3- 3/4+ 3/5+ 1/72- 2/9- 1/36+ 1/15
A= ( 1/3- 3/5+ 1/15) - (3/4- 1/72+ 2/9+ 1/36)
A= (5/15- 9/15+ 1/15) - (54/72- 1/72+ 16/72+ 2/36)
A= 1- 71/72
A= 1/72
 
 

A=1/72

A = \(\frac{1}{3}\)- \(\frac{3}{4}\) - ( - \(\frac{3}{5}\)) + \(\frac{1}{72}\)- \(\frac{2}{9}\)- \(\frac{1}{36}\)+ \(\frac{1}{15}\)

A =   (  \(\frac{1}{3}\)+ \(\frac{3}{5}\)+ \(\frac{1}{15}\)) -  ( \(\frac{3}{4}\)- \(\frac{1}{72}\)+ \(\frac{2}{9}\)+ \(\frac{1}{36}\))

A =    ( \(\frac{5}{15}\)+ \(\frac{9}{15}\)+ \(\frac{1}{15}\)) - ( \(\frac{54}{72}\)- \(\frac{1}{72}\)+ \(\frac{16}{72}\)+ \(\frac{2}{72}\))

A =          1               -            \(\frac{71}{72}\)

A =                  \(\frac{1}{72}\)

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=\(\frac{36-4+3}{6}-\frac{30+10-9}{6}-\frac{18-14+15}{6}\)

=\(\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=\frac{35-31-19}{6}-\frac{15}{6}=-\frac{5}{2}\)

bài này thì dễ:

\(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)

Cách 1:

\(\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=\frac{35-31-19}{6}=-\frac{15}{6}=-\frac{5}{2}\)

Cách 2: 

\(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}=-2\frac{1}{2}=-\frac{5}{2}\)

Cách 1 :

\(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)

\(=\left(\frac{36-4+3}{6}\right)-\left(\frac{30+10-9}{6}\right)-\left(\frac{18-14+15}{8}\right)\) 

\(=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=-\frac{15}{6}=-\frac{5}{2}\)

Cách 2 : \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)

\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)

\(=\left(6-5-3\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)+\left(-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}\right)\)

=\(-2-\frac{1}{2}=-2\frac{1}{2}=-\frac{5}{2}\)

mk làm bài 1 thui,bài 2 chỉ qui đồng ms

3a/6 = 3b/4 => 3(a-b)/ (6-4) = 3.4,5/2= 13,5/2 =k

a = 2k=13,5

b = 4k/3 =9

Ta có: Tử là:

B=\(\frac{1}{2013}+\frac{2}{2012}+...+\frac{2012}{2}+\left(1+1+...+1\right)\)            (2013 số hạng 1)

   =\(\left(\frac{1}{2013}+1\right)+\left(\frac{2}{2012}+1\right)+...+\left(\frac{2012}{2}+1\right)+\left(1\right)\)

  =\(\frac{2014}{2013}+\frac{2014}{2012}+...+\frac{2014}{2}+\frac{2014}{2014}\)

 =\(2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)

=>A=\(\frac{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=2014

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