\(\frac{1}{2x4}\)+ \(\frac{1}{4x6}\)+ ... + \(\frac{1}{98x100}\)= \(\frac{1}{2}\)x(\(\frac{4-2}{2x4}\)+\(\frac{6-4}{4x6}\)+ ... + \(\frac{100-98}{98x100}\))

                                                        = \(\frac{1}{2}\)x(\(\frac{1}{2}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{8}\)+ ... + \(\frac{1}{98}\)-\(\frac{1}{100}\))

                                                        = \(\frac{1}{2}\)x(\(\frac{1}{2}\)-\(\frac{1}{100}\)) = \(\frac{49}{200}\)

kết quả là 49/50

49'/50

\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{96.98}+\frac{1}{98.100}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}\)

\(=\frac{49}{100}\)

bậy bạ

?

><

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

A = \(1-\frac{1}{100}\)

A = \(\frac{100}{100}-\frac{1}{100}\)

A = \(\frac{99}{100}\)

\(A=\frac{1}{1x2}+\frac{1}{2x3}+\frac{ 1}{3x4}+...+\frac{1}{99x100}\)

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

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

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

1/2.4 + 1/4.6 + 1/6.8 + ... + 1/96.98 + 1/98.100

= 1/2.(2/2.4 + 2/4.6 + 2/6.8 + ... + 2/96.98 + 2/98.100)

= 1/2.(1/2 - 1/4 + 1/4 - 1/6 + ... + 1/96 - 1/98 + 1/98 - 1/100)

= 1/2.(1/2 - 1/100)

= 1/2.49/100

= 49/200

49/200

duyệt đi

kết quả bằng 49/200

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