Ta có \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{9}{10}\)

Ta có \(B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=\frac{1}{2}-\frac{1}{7}\)

\(=\frac{5}{14}\)

Ta có \(C=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{1}{6}-\frac{1}{22}\)

\(=\frac{4}{33}\)

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

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

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

\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(B=\frac{1}{2}-\frac{1}{7}\)

\(B=\frac{5}{14}\)

\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)

\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}\right)\)

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(C=\frac{1}{6}-\frac{1}{22}=\frac{4}{33}\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{9\cdot10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}=\frac{9}{10}\)

A=1/3.5+1/5.7+1/7.9+...+1/99.101

2A= 2/3.5+2/5.7+2/7.9+...+2/99.101

2A= 1/3-1/5+1/5-1/7-1/7+1/7-1/9+...+1/99-1/101

2A=1/3-1/101=98/303

A=(98/303)/2=49/303

A=97

   666

A=1/3.5 + 1/5.7 + 1/7.9 +...+ 1/99.101

=1/2.[(1/3-1/5) + (1/5-1/7) + ... + 1/99-1/101)]

=1/2.(1/3-1/101)

=49/303

\(A=1/3.5+1/5.7+1/7.9+…+1/99.101\)

A.2=2/3.5+2/5.7+2/7.9+…+2/99.101

A.2=1/3-1/5+1/5-1/7+1/7-1/9+...+1/99-1/101

Vậy

A.2=1/3-1/101=98/303

A=98/303/2=49/303

Đúng không

A = 1/15 + 1/35 + 1/63 + 1/99 + ... + 1/9999

   = 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 + ... + 1/99x101

A x 2 = 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + ... + 2/99x101

         = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + ... + 1/99 - 1/101

         = 1/3 - 1/101 = 98/303

Vậy A = 98/303 : 2 = 49/303

B = \(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{15}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{63}\)

B = \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{12}+\frac{1}{15}+\frac{1}{20}\right)+\left(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)+\frac{1}{63}\)

B = \(1+\frac{1}{5}+\frac{3}{40}+\frac{1}{63}\)

B = \(1\frac{11}{40}+\frac{1}{63}\)

B = \(1\frac{733}{2520}\)

nguyentuantai làm hòa với Nguyễn Đình Dũng phải chăng mục đích là lấy **** ko

A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với

a)\(\frac{1}{2}-2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{48.50}\right)\)

=\(\frac{1}{2}-\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+.....+\frac{2}{48.50}\right)\)

=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{48}-\frac{1}{50}\right)\)

=\(\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{50}\right)\)

=\(\frac{1}{50}\)

\(1)a)\frac{1}{2}-2\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\right)\)

\(=\frac{1}{2}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{24.25}\right)\)

\(=\frac{1}{2}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{24}-\frac{1}{25}\right)\)

\(=\frac{1}{2}-\left(1-\frac{1}{25}\right)\)

\(=\frac{1}{2}-\frac{24}{25}=\frac{-23}{50}\)

\(\)

\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(A=\frac{1}{2}.\frac{98}{303}\)

\(A=\frac{49}{303}\)

A= \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)

2A=\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

2A=\(\frac{1}{3}-\frac{1}{101}\)

2A=\(\frac{98}{303}\)

A=\(\frac{98}{303}.\frac{1}{2}\)

A=\(\frac{49}{303}\)

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