\(A=\frac{2}{2\cdot5}+\frac{2}{5\cdot8}+\frac{2}{8\cdot11}+...+\frac{2}{92\cdot95}+\frac{2}{95\cdot98}\)

\(A=\frac{2}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\right]\)

\(A=\frac{2}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right]\)

\(A=\frac{2}{3}\left[\frac{1}{2}-\frac{1}{98}\right]=\frac{2}{3}\left[\frac{49}{98}-\frac{1}{98}\right]=\frac{2}{3}\cdot\frac{48}{98}=\frac{2}{3}\cdot\frac{24}{49}=\frac{2}{1}\cdot\frac{8}{49}=\frac{16}{49}\)

\(A=\frac{2}{2.5}+\frac{2}{5.8}+...+\frac{2}{92.95}+\frac{2}{95.98}\)

\(=\frac{2}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)

\(=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)

\(=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(=\frac{2}{3}.\frac{24}{49}\)

\(=\frac{16}{49}\)

=1/3(3/2*5+3/5*8+...+3/95*98)

=1/3(1/2-1/5+1/5-1/8+...+1/95-1/98)

=1/3*96/196

=32/196

=8/49

Ta có:\(A=\dfrac{1}{2}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{4}{11}+...+\dfrac{31}{92}-\dfrac{32}{95}+\dfrac{32}{95}-\dfrac{33}{98}\)

                \(=\dfrac{1}{2}+\dfrac{33}{98}=\dfrac{82}{98}=\dfrac{41}{49}\)

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{95.98}\)

=> 3A = \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\)

=> 3A = \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\)

=> 3A = \(\frac{1}{2}-\frac{1}{98}\)

=> 3A = \(\frac{24}{49}\)

=> A = \(\frac{8}{49}\)

\(A=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\)

\(A=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\right)\)

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)=\frac{1}{3}\cdot\frac{24}{49}=\frac{8}{49}\)

Sửa 95.98 thành 1/(95.98) nhá

đặt

\(A=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+..+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot97}\)

\(3A=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{92\cdot95}+\dfrac{3}{95\cdot97}\)

\(3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{97}\)

\(3A=\dfrac{1}{2}-\dfrac{1}{97}\\ 3A=\dfrac{95}{194}\\ A=\dfrac{95}{582}\)

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

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

     \(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}\)

   \(=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{92-92}{92.95}\)

   \(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\)

     \(=\frac{1}{2}-\frac{1}{95}=\frac{93}{190}\)

\(C=\frac{5}{6}+\frac{5}{66}+\frac{5}{176}+\frac{5}{336}\)

    \(=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}\)

    \(=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+\frac{21-16}{16.21}\) 

    \(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}\)

     \(=1-\frac{1}{21}=\frac{20}{21}\)

[ HỌC TỐT]

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

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

\(A=\frac{1}{2}-\frac{1}{100}\)

\(A=\frac{100}{200}-\frac{2}{200}\)

\(A=\frac{98}{200}=\frac{49}{100}\)

a) 19/23

b) -5/11

a) 8/3 x 2/5 x 3/8 x 10 x 19/92

= ( 8/3 x 3/8 ) x ( 2/5 x 10 ) x 19/92

=     1 x 4 x 19/92

=     4 x 19/92   =   76/92  =  19/23

b) N = 5/7 x 5/11 + 5/7 x 2/11 - 5/7 x 14/11

       = 5/7 x ( 5/11 + 2/11 - 14/11 )

       = 5/7 x ( -7/11 )

       =   -5/11

9992 + 185 =122222

94/95 nhé bạn