1/2 + 1/6 + 1/12 + ... + 1/9900 + 1/10100

= 1/1.2 + 1/2.3 + 1/3.4 +... +1/99.100 + 1/100.101

= 1/1 - 1/2 + 1/2 + 1/3 - 1/3 + 1/4 +... + 1/99 - 1 / 100 + 1/100 - 1/101

= 1/1 - 1/101

= 100 /101

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{9900}+\frac{1}{10100}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}+\frac{1}{100.101}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

=\(1-\frac{1}{101}\)

=\(\frac{100}{101}\)

Mik trả lời ở bài dưới rồi đó.

o dau ha ho thu giang

Ta có: \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}+\frac{1}{10100}\)

     \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}+\frac{1}{100.101}\)

     \(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

     \(=1-\frac{1}{101}\)

     \(=\frac{100}{101}\)

Tương đương \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}+\frac{1}{100.101}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)

= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +....+1 /99.100

= 1/1 - 1/2 + 1/2 -1/3 + .... + 1/99 - 1/100

= 1/1 - 1/100

= 100/100 - 1/100

= 99/100

1/2+1/6+1/12+1/20+...+1/9900

=1/1.2+1/2.3+1/3.4+...+1/99.100

=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

=1-1/100=99/100

1/2+1/6+1/12+...+1/9900

=1/1.2+1/2.3+1/3.4+...+1/99.100

=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

=1-1/100=99/100

1/2 + 1/6 + 1/12 + 1/20 + ... + 1/9900

= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/99.100

= 1 - 1/2 + 1/2 - 1/2 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100

= 1 - 1/100

= 99/100 

Mk nhanh nhất đó

Đúng 100%

Tk mk mk tk lại

Cảm ơn bạn nhiều

Thank you very much

( ^ _ ^ )

99/100

Buổi chiều hôm nay cô giáo mới dạy cho mình mà nên mình chắc chắn 100%

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{99\cdot100}\)

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

\(A=\dfrac{1}{1}-\dfrac{1}{100}\)

\(A=\dfrac{99}{100}\)

\(\cdot\) LÀ DẤU \(\times\)

A = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+.....+ \(\dfrac{1}{9900}\)

A = \(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+....+\dfrac{1}{99\times100}\)

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

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)

A = \(\dfrac{99}{100}\)

49/100

\(\left(1-\frac{2}{6}\right)\left(1-\frac{2}{12}\right)...\left(1-\frac{2}{9900}\right)\)

\(=\frac{4}{6}.\frac{10}{12}...\frac{9898}{9900}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{98.101}{99.100}\)

\(=\frac{1.2...98}{3.4...100}.\frac{4.5...101}{2.3...99}\)

\(=\frac{2}{99.100}.\frac{100.101}{2.3}\)

\(=\frac{101}{99.3}\)

\(=\frac{101}{297}\)

đáp số:\(\frac{101}{297}\)

ai k mk mk sẽ k lại ^-^

là 101/247

l

1/2+1/6+1/12+...+1/9900
=1/(1*2)+1/(2*3)+1/(3*4)+...+1/(99*100)
=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-1/100

=99/100

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

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

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)