\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}=\frac{5}{6}\)

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

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

\(=\dfrac{1}{1}\cdot\dfrac{1}{2}+\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{1}{7}\)

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

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

Ta có: \(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9702}\)

\(=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{98\cdot99}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}\)

\(=\dfrac{32}{99}\)

Giải:

1/12+1/20+1/30+...+1/9702

=1/3.4+1/4.5+1/5.6+...+1/98.99

=1/3-1/4+1/4-1/5+1/5-1/6+...+1/98-1/99

=1/3-1/99

=32/99

Chúc bạn học tốt!

Ta sử dụng t/c sau:

`1/(a(a+1))=1/a-1/(a+1)`

`=>1/12+1/20+1/30+...+1/9072`

`=1/(3.4)+1/(4.5)+....+1/(98.99)`

`=1/3-1/4+1/4-1/5+....+1/98-1/99`

`=1/3-1/99`

`=32/99`

 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 
=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/(9.10) 
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5.+1/5-1/6... 
+1/9-1/10 
=1-1/10 
=9/10

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

=1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8

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

=1/2-1/8

=3/8

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

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

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

\(=\frac{1}{2}-\frac{1}{8}\)

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

A=1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7=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=6/7

=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)+(1/2-1/2)+(1/3-1/3) + ... +(1/6-1/6)

= (1-1/7) + 0+...+0 = 1-1/7 = 7/7 - 1/7 = 6/7

Vậy A =6/7

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

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

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

k cho mk nha

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

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

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

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

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

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