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cái này tính cái gì thế

ko hiểu

\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\)

\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)

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

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

\(=\frac{25-1}{100}=\frac{24}{100}=\frac{6}{25}\)

\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{9900}\)

\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)

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

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

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

k mình nha !

=1/4*5+1/5*6+1/6*7+...+1/99*100.

=1/4-1/5+1/5-1/6+1/6-1/7+...+1/99-1/100.

=1/4-1/100=6/25.

k nha có j kb

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)

\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

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

\(=\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\)

Vậy...

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\)

\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

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

\(=\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\left(đpcm\right)\)

Vậy...

Giải 

\(A=1+2+3+4+5+...+99+100\)

Số số hạng của A là: \(\left(100-1\right)\div1+1=100\)(số hạng)

Tổng A là: \(\frac{\left(100+1\right)\times100}{2}=5050\)

Vây A=5050

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

\(B=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)

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

\(B=1-\frac{1}{100}=\frac{99}{100}\)

Vậy \(B=\frac{99}{100}\)

minh cam thay de hoi sai

A = 1 + 2 + 3 + ... + 99 + 100

A = 100 + 99 + ... + 2 + 1

2A = 101 + 101 +... + 101 + 101 ( 100 số hạng )

A = 101 . 100 : 2 = 5050

Vậy A = 5050 

B = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + ... + 1/9900

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

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

B = 1/1 - 1/100

B = 99/100

Vậy B = 99/100

\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)

\(A=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{99\cdot100}\)

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

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

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

\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{9900}\)

\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+.....+\frac{1}{99.100}\)

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

\(=\frac{1}{4}-\frac{1}{100}\\ =\frac{24}{100}=\frac{6}{25}\)

= 1-1/2.3-1/3.4-....-1/99.100

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

= 1-1/2+1/100

= 51/100

Tk mk nha

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

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

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

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

\(\cdot\) 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