\(E=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{240}\)

\(\Rightarrow E=\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{15.16}\)

\(E=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(E=2.\frac{3}{16}=\frac{3}{8}\)

\(E=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)

\(E=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{15.16}\)

\(E=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(E=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)

\(E=2.\frac{3}{16}\)

\(E=\frac{3}{8}\)

\(D=\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+...+\frac{1}{336}\)

\(D=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{16.21}\)

\(D=\frac{1}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{16}-\frac{1}{21}\right)\)

\(D=\frac{1}{5}.\left(1-\frac{1}{21}\right)\)

\(D=\frac{1}{5}.\frac{20}{21}\)

\(D=\frac{4}{21}\)

\(A=\frac{4}{2}+\frac{4}{6}+\frac{4}{12}+\frac{4}{20}+\frac{4}{30}+\frac{4}{42}\)

\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+\frac{4}{4.5}+\frac{4}{5.6}+\frac{4}{6.7}\)

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

\(A=4\left(1-\frac{1}{7}\right)\)

\(A=4.\frac{6}{7}\)

\(A=\frac{24}{7}\)

\(A=\frac{4}{2}+\frac{4}{6}+\frac{4}{12}+\frac{4}{20}+\frac{4}{30}+\frac{4}{42}=4\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

\(=4\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

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

\(=4\left(1-\frac{1}{7}\right)=\frac{6}{7}.4=\frac{24}{7}\)

A = \(\frac{4}{2}+\frac{4}{6}+\frac{4}{12}+\frac{4}{20}+\frac{4}{30}+\frac{4}{42}\)

A = \(\frac{4}{2}+\frac{4}{2.3}+\frac{4}{3.4}+\frac{4}{4.5}+\frac{4}{5.6}+\frac{4}{6.7}\)

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

A x 4 = \(1-\frac{1}{7}\)

A x 4 = \(\frac{6}{7}\)

      A = \(\frac{6}{7}:4\)

      A = \(\frac{3}{14}\)

\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)

\(A=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{41}+1\)\(-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)

\(A=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(A=9-\left[1\div\left(1\times2\right)+1\div\left(2\times3\right)+1\div\left(3\times4\right)+1\div\left(4\times5\right)\right]\)\(+1\div\left(5\times6\right)+1\div\left(6\times7\right)+1\div\left(7\times8\right)+1\div\left(8\times9\right)\)\(+1\div\left(9\times10\right)\)]

\(A=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)\(+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\))

\(A=9-\left(1-\frac{1}{10}\right)\)

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

\(A=\frac{81}{10}\)

\(A=1+1+...+1-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)

\(A=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)

\(A=9-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{10-9}{10.9}\right)\)

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

\(A=9-\frac{9}{10}=\frac{81}{10}\)

cho hỏi viết phân số kiểu gì???????

\(B1\)

\(=\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{37}-\frac{1}{38}-\frac{1}{39}\)

\(=1-\frac{1}{39}\)

\(=\frac{38}{39}\)

\(B2\)

\(=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+.....+\frac{1}{99\cdot100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+......+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}\)

\(=\frac{25}{100}-\frac{1}{100}\)

\(=\frac{24}{100}\)

\(=\frac{6}{25}\)

Bài 1 :

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)

\(=\frac{1}{1.2}-\frac{1}{38.39}\)

\(=\frac{370}{741}\)

Tớ chỉ biết làm bài 1 thui

1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/37.38.39

= 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 - 1/3.4 + 1/3.4 - 1/4.5 + 1/37.38 - 1/38.39

= 1/1.2 - 1/38.39

= 370/741

\(M=\left(\dfrac{3}{2}-\dfrac{5}{6}+\dfrac{7}{12}-\dfrac{17}{72}\right)+\left(-\dfrac{9}{20}+\dfrac{11}{30}\right)+\left(\dfrac{-13}{42}+\dfrac{15}{56}\right)\)

\(=\dfrac{108}{72}-\dfrac{60}{72}+\dfrac{42}{72}-\dfrac{17}{72}+\dfrac{-27}{60}+\dfrac{22}{60}+\dfrac{-52}{168}+\dfrac{45}{168}\)

\(=\dfrac{73}{72}-\dfrac{1}{12}-\dfrac{1}{24}=\dfrac{73}{72}-\dfrac{6}{72}-\dfrac{3}{72}=\dfrac{64}{72}=\dfrac{8}{9}\)

1/

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

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+1/7-1/9+ ... + 1/99-1/101

2A=1/3-1/101

A=49/303

Sai thì thôi nhé

A= 1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7

A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7

A=1-1/7

A=6/7

A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)

\(A=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)\)

\(A=1-\frac{1}{9}=\frac{8}{9}\)

A=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)

=1\(-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

=1-\(\frac{1}{9}=\frac{8}{9}\)

Vậy A=\(\frac{8}{9}\)