Ta có: \(A=-\frac{4}{20}-\frac{4}{30}-\frac{4}{42}-\frac{4}{56}-\frac{4}{72}-\frac{4}{90}\)

   \(\Leftrightarrow A=-4.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

   \(\Leftrightarrow A=-4.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

   \(\Leftrightarrow A=-4.\left(\frac{1}{4}-\frac{1}{10}\right)\)

   \(\Leftrightarrow A=-4.\left(\frac{5-2}{20}\right)\)

   \(\Leftrightarrow A=-4.\frac{3}{20}=-\frac{3}{5}\)

Vậy \(\Leftrightarrow A=-\frac{3}{5}\)

thanks

A=-4/4.5+-4/5.6+-4/6.7+-4/7.8+-4/8.9+-4/9.10

-1/4A=1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10

-1/4A=1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10

-1/4A=1/4-1/10

-1/4A=3/20

A=-3/5

thanks

Ta có : A = -4/20+-4/30+-4/42+-4/56+-4/72+-4/90

                = -4/4.5+-4/5.6+-4/6.7 +-4/7.8+-4/8.9+-4/9.10

                 =-4/4-(-4/5)+(-4/5)-(-4/6)+......+(-4/9)-(-4/10)

                  =-4/4-(-4/10)

                 =-3/5

# chúc bạn học tốt #

\(A=\frac{-4}{20}+\frac{-4}{30}+\frac{-4}{42}+...+\frac{-4}{90}\)

     =\(-4\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\right)\)

    =-4\(\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{9.10}\right)\)

  =-4\(\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}\right)\)

=-4\(\left(\frac{1}{4}-\frac{1}{10}\right)\)

=-4.\(\frac{3}{20}\)=\(\frac{-3}{5}\)

a,58,25974026

b, 7/15

a) \(\frac{A}{7}=\frac{5}{2\times7}+\frac{4}{7\times11}+\frac{3}{11\times14}+\frac{1}{14\times15}+\frac{13}{15\times28}\)

\(\frac{A}{7}=\frac{7-2}{2\times7}+\frac{11-4}{7\times11}+\frac{14-11}{11\times14}+\frac{15-14}{14\times15}+\frac{28-15}{15\times28}\)

\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)

\(A=7.\frac{13}{28}=\frac{13}{4}\)

ban loan dung

a) \(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

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

b) \(=\dfrac{1}{3x4}+\dfrac{1}{4x5}+\dfrac{1}{5x6}+\dfrac{1}{6x7}+\dfrac{1}{7x8}+\dfrac{1}{8x9}+\dfrac{1}{9x10}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{1}{3}-\dfrac{1}{10}=\dfrac{10}{30}-\dfrac{3}{30}=\dfrac{7}{30}\)

Ko biết ahihihi

a)A=1/20+1/30+1/42+1/56+1/72+1/90+1/110

= 1/4*5 + 1/5*6 + 1/6*7 + ... + 1/10*11

= 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/10 - 1/11 

= 1/4 - 1/11

= 7/44

b)B=1/2+1/4+1/6+1/8+...+1/512+1/1024

B = 1/2^1 + 1/2^2 + 1/2^3 + ... + 1/2^9 + 1/2^10

2B = 1 + 1/2 + 1/2^2 + ... + 1/2^10 + 1/2^11

2B - B = B = 1 + 1/2^11

a) Ta có: \(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)

        \(\Leftrightarrow A=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)

        \(\Leftrightarrow A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

        \(\Leftrightarrow A=\frac{1}{4}-\frac{1}{11}\)

        \(\Leftrightarrow A=\frac{11-4}{44}\)

        \(\Leftrightarrow A=\frac{7}{44}\)

a. \(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+......+\dfrac{3}{17.20}\)

\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+......+\dfrac{1}{17}-\dfrac{1}{20}\)

\(=\dfrac{1}{2}-\dfrac{1}{20}\)

\(=\dfrac{9}{20}\)

b. \(B=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{1}{4}-\dfrac{1}{10}\)

\(=\dfrac{3}{20}\)

c. \(C=\dfrac{4^2}{1.5}+\dfrac{4^2}{5.9}+......+\dfrac{4^2}{45.49}\)

\(=4\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+....+\dfrac{4}{45.49}\right)\)

\(=4\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+.....+\dfrac{1}{45}-\dfrac{1}{49}\right)\)

\(=4\left(1-\dfrac{1}{49}\right)\)

\(=4.\dfrac{48}{49}\)

\(=\dfrac{192}{49}\)