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

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

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

cho mình nha!

=1/2 - 1/4 + 1/4 - 1/6 + ... + 1/98 - 1/100

=1/2 - 1/100 = 49/100

1/2 - 1/4 +  1/4 - 1/6 + 1/6 - 1/8 + ... + 1/96 - 1/98 + 1/98 - 1/100

= 1/2 - 1/100 

= 49/100

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 + 1/6 - 1/8 + ... + 1/96 - 1/98 + 1/98 - 1/100 )

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

= 1/2 . ( 50/100 - 1/100 )

= 49/200

99/200

\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+....+\frac{1}{98\cdot100}\)

\(=\frac{1}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+.......+\frac{2}{98\cdot100}\right)\)

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

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

Đặt A = 1/2.4 + 1/4.6 + .... +1/98.100

\(2A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\)

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

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

         \(=\frac{49}{100}\)

\(\Rightarrow A=\frac{49}{100}\div2=\frac{49}{200}\)

M = \(\dfrac{5}{2.4}\) + \(\dfrac{5}{4.6}\)+ \(\dfrac{5}{6.8}\)+ ...+ \(\dfrac{5}{96.98}\)+ \(\dfrac{5}{98.100}\)

M = \(\dfrac{5}{2}\).( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+ \(\dfrac{2}{96.98}\)+ \(\dfrac{2}{98.100}\))

M = \(\dfrac{5}{2}\).( \(\dfrac{1}{2}-\dfrac{1}{4}\)+ \(\dfrac{1}{4}-\dfrac{1}{6}\)+ \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\)+...+ \(\dfrac{1}{96}\)-\(\dfrac{1}{98}\)+ \(\dfrac{1}{98}\)-\(\dfrac{1}{100}\))

M = \(\dfrac{5}{2}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))

M = \(\dfrac{49}{40}\)

\(x\) \(\times\) M - 1 = \(\dfrac{20}{29}\)

\(x\) \(\times\) \(\dfrac{49}{40}\) = \(\dfrac{20}{29}\) + 1

\(x\) \(\times\) \(\dfrac{49}{40}\) = \(\dfrac{49}{29}\)

\(x\)           = \(\dfrac{49}{29}\) : \(\dfrac{49}{40}\)

\(x\)           = \(\dfrac{40}{29}\)

=1/2-1/4+1/4-1/6+....+1/98-1/100

=1/2-1/100

=49/100

=1/2-1/4+1/4-1/6+ ... +1/98-1/100

=1/2-1/100

=49/100

C= \(\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{98.100}\right)\)

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

  = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)

  = \(\frac{1}{2}.\left(\frac{50}{100}-\frac{1}{100}\right)\)

  = \(\frac{1}{2}.\frac{49}{100}\)

  = \(\frac{49}{200}\)

\(\frac{2.4+4.6+6.8+...+98.100}{1.2+2.3+3.4+...+49.50}=\frac{4.\left(1.2+2.3+3.4+...+49.50\right)}{1.2+2.3+3.4+...+49.50}=\frac{4}{1}=4\)

(0,5+0,5)÷0,5_0,5×0,5=

\(\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)

= \(\frac{5}{2}-\frac{5}{4}+\frac{5}{4}-\frac{5}{6}+...+\frac{5}{98}-\frac{5}{100}\)

= \(\frac{5}{2}-\frac{5}{100}\)

= \(\frac{49}{50}\)

\(Q=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)

    \(=5\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

    \(=\frac{5}{2}.2.\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

    \(=\frac{5}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)

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

    \(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{5}{2}.\frac{49}{100}=\frac{49}{40}\)

\(\Rightarrow Q=\frac{49}{40}\)

(20182018 . 2019 - 20192019.2018 ) . (2.4 + 4.6 + 6.8 +...+ 98.100 )

=(2018.10001 . 2019 - 20192019.2018 ) . (2.4 + 4.6 + 6.8 +...+ 98.100 )

=(2018.20192019-20192019.2018). (2.4 + 4.6 + 6.8 +...+ 98.100 )

=0. (2.4 + 4.6 + 6.8 +...+ 98.100 )

=0