\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)...\times\left(1-\frac{1}{100}\right)\)

=\(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times...\times\frac{99}{100}\)

=\(\frac{1\times2\times3\times4\times...\times99}{2\times3\times4\times5\times...\times100}\)

= \(\frac{1}{100}\)

A=1/2+1/6+....+1/56+1/72

A=1/1.2+1/2.3+...+1/7.8+1/8.9

A=1/1-1/2+1/2-1/3+...+1/7-1/8+1/8-1/9

A=1/1-1/9=9/9-1/9=8/9

Thanks ban nha

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

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

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

\(\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+\frac{1}{9x11}\right)xX=\frac{8}{11}\)

\(\Rightarrow\left(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right)xX=\frac{16}{11}\)

\(\Rightarrow\left(\frac{5-3}{3x5}+\frac{7-5}{5x7}+\frac{9-7}{7x9}+\frac{11-9}{9x11}\right)xX=\frac{16}{11}\)

\(\Rightarrow\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)xX=\frac{16}{11}\)

\(\Rightarrow\left(\frac{1}{3}-\frac{1}{11}\right)xX=\frac{16}{11}\Rightarrow\frac{8}{33}xX=\frac{16}{11}\)

\(\Rightarrow X=\frac{16}{11}:\frac{8}{33}=\frac{16}{11}x\frac{33}{8}=6\)

\(C=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{97.98}+\frac{1}{99.100}\)

\(C=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{99}-\frac{1}{100}\)

\(C=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{98}+\frac{1}{100}\right)\)

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

\(C=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)

\(C=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(D=\frac{1}{51.100}+\frac{1}{52.99}+\frac{1}{53.98}+...+\frac{1}{99.52}+\frac{1}{100.51}\)

\(D=\frac{1}{151}.\left(\frac{151}{51.100}+\frac{151}{52.99}+\frac{151}{53.98}+...+\frac{151}{99.52}+\frac{151}{100.51}\right)\)

\(D=\frac{1}{151}.\left(\frac{1}{100}+\frac{1}{51}+\frac{1}{99}+\frac{1}{52}+...+\frac{1}{52}+\frac{1}{99}+\frac{1}{51}+\frac{1}{100}\right)\)

\(D=\frac{1}{151}.\left(\frac{2}{100}+\frac{2}{99}+...+\frac{2}{51}\right)\)

\(D=\frac{2}{151}.\left(\frac{1}{100}+\frac{1}{99}+...+\frac{1}{51}\right)\)

\(\Rightarrow C:D=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{2}{151}.\left(\frac{1}{100}+\frac{1}{99}+...+\frac{1}{51}\right)}\)

\(\Rightarrow C:D=\frac{151}{2}=75\frac{1}{2}\)

Khó hiểu vậy ạ, giảng kĩ đc ko bạn :)

bạn giải thích kĩ đực không? Khó hiểu quá!

E=1/1.3+1/3.6+1/6.9+.............+1/20.23

<=>E=1/1-1/3+1/3-1/6+1/6-1/9+...........+1/20-1/23

<=>E=1/1-1/23

<=>E=22/23

Kb và k mk nha mn.

Cho hai phan so 1/n va 1/n+1 (n thuoc z)chung to rang h cua hai phan so nay bang hieu cua chung

\(=\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot....\cdot\left(1+\frac{1}{2010}\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{2011}{2010}\)

\(=\frac{3\cdot4\cdot5\cdot...\cdot2011}{2\cdot3\cdot4\cdot...\cdot2010}\)

\(=\frac{2011}{2}\)

Chúc em học tốt

=10

tk nha

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 × 0

= 10

Tk nha

Ai ko nhan tin mk tk 1 tk thui nha