\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(< =>\frac{128}{256}+\frac{64}{256}+\frac{32}{256}+\frac{16}{256}+\frac{8}{256}+\frac{4}{256}+\frac{2}{256}+\frac{1}{256}\)

\(< =>\frac{128+64+32+16+8+4+2+1}{256}\)

\(< =>\frac{255}{256}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

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

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

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

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

\(< =>\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)

\(< =>\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\)

\(< =>\frac{1}{100}\)

mk chuc ban hoc tot nhe :))

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

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

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

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

Cho hai số biết rằng bớt số thứ nhất 28 đơn vị thì được số thứ hai va 1/3 số thứ nhất bằng 3/5 số thứ hai.Tìm hai số đó

sai roi

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

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

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

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

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

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1}-\frac{1}{100}\)
\(\frac{100-1}{100}\)
\(\frac{99}{100}\)

 

99/100

Đặt A = 1/1x2 + 1/2x3 + 1/3x4 + .... + 1/99x100

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

=> A = 1 - 1/100

=> A = 99/100

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

hok

tốt

nha

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

=1-1/100=99/100

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

= 1 - 1/100

= 99/100

= 99/100

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{100\cdot101}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{100}-\dfrac{1}{101}\)

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

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

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

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

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

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

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

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

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

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

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

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