Bài làm 

\(D=\frac{6}{3,5}+\frac{6}{5.7}+...+\frac{6}{21.23}\)

\(D=3.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{21.23}\right)\)

\(D=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)

\(D=3.\left(\frac{1}{3}-\frac{1}{23}\right)\)

\(D=3.\frac{20}{69}\)

\(D=\frac{20}{23}\)

Học tốt 

Bài làm 

 \(D=\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{21.23}\)

\(D=3.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{21.23}\right)\)

\(D=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)

\(D=3.\left(\frac{1}{3}-\frac{1}{23}\right)\)

\(D=3.\frac{20}{69}\)

\(D=\frac{20}{23}\)

   \(E=\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\)

\(E=10.\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)\)

\(E=10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\)

\(E=10.\left(\frac{1}{11}-\frac{1}{55}\right)\)

\(E=10.\frac{4}{55}\)

\(E=\frac{8}{11}\)

     \(G=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)

\(G=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(G=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

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

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

Nhớ k cho m nha 

Ta có:

\(D=\frac{6}{3\cdot5}+\frac{6}{5\cdot7}+...+\frac{6}{21\cdot23}\)

\(D=3.\left(\frac{2}{3.5}+\frac{2}{5\cdot7}+...+\frac{2}{21.23}\right)\)

\(D=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)

\(D=3.\left(\frac{1}{3}-\frac{1}{23}\right)\)

\(D=3.\frac{20}{69}\)

\(D=\frac{20}{23}\)

Vậy \(D=\frac{20}{23}\)

= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9

= 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/9

=1-1/9

=8/9

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

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

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

= \(1-\frac{1}{9}\)

= \(\frac{8}{9}\)

1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72

= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9

= 1 - 1/9

= 8/9

Đúng rùi bn

BN mún hỏi j vậy, đây k phải câu hỏi, mà có thì phải là toán lớp 6

A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{72}+\frac{1}{81}\)

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{8\times9}+\frac{1}{81}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{81}\)

\(A=1-\frac{1}{9}+\frac{1}{81}=\frac{73}{81}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{81}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{81}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{81}\)

\(=1-\frac{1}{9}+\frac{1}{81}\)

\(=\frac{8}{9}+\frac{1}{81}\)

\(=\frac{73}{81}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{8\cdot9}+\frac{1}{81}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{81}\)
\(=1-\frac{1}{9}+\frac{1}{81}\)
\(=\frac{8}{9}+\frac{1}{81}\)
\(=\frac{73}{81}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)

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

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

\(=\frac{6}{7}\)

1/2+1/6+1/12+1/20+1/30+1/42

=1/1×2+1/2×3+1/3×4+1/4×5+1/5×6+1/6×7

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

=1-1/7

=6/7.

Tích mk nha

1/2+1/6+1/12+1/20+1/30+1/42

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

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

=1-1/7

=6/7

1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 

= 1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/6x7 

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

= 1/1 - 1/7 

= 7/7 - 1/7 

= 6/7 

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

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

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

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

\(=\frac{6}{7}\)

thank you ~! >-0

\(=\frac{1}{2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}\)

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

\(=1-\frac{1}{7}=\frac{7-1}{7}=\frac{6}{7}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

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

\(=\frac{1}{1}+0+0+.....+0-\frac{1}{7}=1-\frac{1}{7}=\frac{6}{7}\)

Lưu ý: Dãy tổng đặc biệt 

\(1-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

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

\(=1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)

\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)

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

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

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

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