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2 tháng 7 2016

\(D=\frac{6}{2x4}+\frac{6}{4x6}+\frac{6}{6x8}+....+\frac{6}{48x50}\)

\(=\frac{6}{2}x\left(\frac{2}{2x4}+\frac{2}{4x6}+\frac{2}{6x8}+....+\frac{2}{48x50}\right)\)

\(=3x\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{48}-\frac{1}{50}\right)\)

\(=3x\left(\frac{1}{2}-\frac{1}{50}\right)=3x\frac{12}{25}=\frac{36}{25}\)

Vậy D=36/25

2 tháng 7 2016

D=6/2x4 + 6/4x6 + 6/6x8 + ...+ 6 /48 x50

D=3 x (2/2x4 + 2/4x6 +  2/6x8 + ...+ 2 /48 x50)

D= 3x (1/2 - 1/4 + 1/4 - 1/6 + 1/6-1/8 + ... + 1/48 - 1/50)

D= 3 x (1/2 - 1/50)

D= 3 x 12/25

D= 36/25

5 tháng 5 2019

\(\frac{6}{2x4}+\frac{6}{4x6}+\frac{6}{6x8}+...+\frac{6}{98x100}\)

\(=3x\left(\frac{2}{2x4}+\frac{2}{4x6}+\frac{2}{6x8}+...+\frac{2}{98x100}\right)\)

\(=3x\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(=3x\left(\frac{1}{2}-\frac{1}{100}\right)=3x\left(\frac{50}{100}-\frac{1}{100}\right)=3x\frac{49}{100}\)

\(=\frac{147}{100}\)

5 tháng 5 2019

\(\frac{6}{2\cdot4}+\frac{6}{4\cdot6}+\frac{6}{6\cdot8}+...+\frac{6}{98\cdot100}\)

=\(\frac{3\cdot2}{2\cdot4}+\frac{3\cdot2}{4\cdot6}+\frac{3\cdot2}{6\cdot8}+...+\frac{3\cdot2}{98\cdot100}\)

=\(\text{​​}3\cdot\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{98\cdot100}\right)\)

=\(\text{​​}3\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)

=\(\text{​​}3\cdot\left(\frac{1}{2}-\frac{1}{100}\right)\)

=\(\text{​​}3\cdot\frac{49}{100}=\frac{147}{100}\)

6 tháng 8 2017

\(a,\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+....+\frac{4}{16.18}+\frac{4}{18.20}\)

\(=\frac{4}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{18}-\frac{1}{20}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=2.\frac{9}{20}\)

\(=\frac{9}{10}\)

\(b,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{9}{10}\)

thank you

8 tháng 7 2015

2[1/2X4+1/4X6+1/6X8+...+1/Xx(X+2)]=11/45x2

2/2x4+2/4x6+2/6x8+....+2/Xx(X+2)=22/45

1/2-1/4+1/4-1/6+1/6-1/8+...+1/x-1/x+2=22/45

1/2-1/x+2=22/45

1/x+2=1/2-22/45

1/x+2=1/90

=>x+2=90

=>x=88

vậy x=88

\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{x\left(x+2\right)}=\frac{11}{45}\)

\(\Rightarrow\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{x\left(x+2\right)}=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+2}=\frac{22}{45}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{2}-\frac{22}{45}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{90}\)

=>x+2=90

=>x=90-2

=>x=88

vậy x=88

26 tháng 6 2023

Em cần phần nào nhỉ .

26 tháng 6 2023

A = \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+\(\dfrac{5}{16.21}\)+...+\(\dfrac{5}{101.106}\)

A = \(\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{106}\)

A = \(\dfrac{105}{106}\)

B = \(\dfrac{3}{1.4}\) +\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)

B = \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\)

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

B = \(\dfrac{99}{100}\)

C = \(\dfrac{1}{2.7}+\dfrac{1}{7.12}\) + \(\dfrac{1}{12.17}\)+...+ \(\dfrac{1}{97.102}\)

C= \(\dfrac{1}{5}\) \(\times\)\(\dfrac{5}{2.7}+\dfrac{5}{7.12}+\dfrac{5}{12.17}+...+\dfrac{5}{97.102}\))

C = \(\dfrac{1}{5}\)\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{17}\)+...+ \(\dfrac{1}{97}\) - \(\dfrac{1}{102}\))

C = \(\dfrac{1}{5}\) \(\times\)\(\dfrac{1}{2}\) - \(\dfrac{1}{102}\))

C = \(\dfrac{1}{5}\) \(\times\) \(\dfrac{25}{51}\)

C = \(\dfrac{5}{51}\) 

D = \(\dfrac{1}{2}\) +   \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)

D = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+\(\dfrac{1}{7.8}\)\(\dfrac{1}{8.9}\)

D = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)

D = \(\dfrac{1}{1}\) - \(\dfrac{1}{9}\)

D = \(\dfrac{8}{9}\)

E = \(\dfrac{3}{2.4}\)+\(\dfrac{3}{4.6}\)+\(\dfrac{3}{6.8}\)+...+\(\dfrac{3}{98.100}\)

E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)\(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))

E = \(\dfrac{3}{2}\)\(\times\)\(\dfrac{1}{2}\) - \(\dfrac{1}{4}\)\(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\))

E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))

E = \(\dfrac{3}{2}\) \(\times\) \(\dfrac{49}{100}\)

E = \(\dfrac{147}{200}\)

S=(2+98)*(4+6)+...+100+100+102

100*10+....+100+100*102
=224400

6 tháng 9 2023

\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{40.42}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{40}-\dfrac{1}{42}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{42}\right)\)

\(=\dfrac{1}{2}.\dfrac{10}{21}\)

\(=\dfrac{5}{21}\)

\(#Wendy.Dang\)

6 tháng 9 2023

\(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{40\cdot42}\)

\(=\dfrac{1}{2}\cdot\left(2\cdot\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{40\cdot42}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{40\cdot42}\right)\)

\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{40}-\dfrac{1}{42}\right)\)

\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{42}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{41}{42}\)

\(=\dfrac{41}{84}\)