B=1/25.27+1/27.29+1/29.31+......+1/73.75
F=4/2.3+4/4.6+4/6.8+.....+4/2008.2010
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a, \(\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
\(=\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\right)\)
\(=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)
\(=\frac{1}{2}\left(\frac{2}{75}\right)\)
\(=\frac{1}{75}\)
b, \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1004}{2010}\right)\)
\(=2\left(\frac{502}{1005}\right)\)
\(=\frac{1004}{1005}\)
Tk hộ =v
\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}.\frac{2}{75}=\frac{1}{75}\)
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)=2.\left(\frac{1}{2}-\frac{1}{2010}\right)=2.\frac{502}{1005}=\frac{1004}{1005}\)
C=4/2.4+4/4.6+4/6.8+...+4/2008.2010
C = 2 ( 2 / 2.4 + 2/4.6 + 2/6.8 + ...+2/2008.2010)
C = 2 ( 1 - 1/4 + 1/4 - 1/6+1/6 - 1/8 +....+1/2008 - 1/2010 )
C = 2 ( 1 - 1 / 2010 )
C = 2 . 2009/2010
C = 2009 / 1005
Chúc bạn học tốt !
\(\text{Ta có:}\) \(C=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(\Rightarrow\frac{1}{2}C=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2008.2010}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2008}-\frac{1}{2010}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{2}-\frac{1}{2010}\)
\(\Rightarrow\frac{1}{2}C=\frac{502}{1005}\)
\(\Rightarrow C=\frac{502}{1005}:\frac{1}{2}=\frac{1004}{1005}\)
Ta có: \(B=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
\(\Rightarrow2B=\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\)
\(\Rightarrow2B=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+....+\frac{1}{73}-\frac{1}{75}\)
\(\Rightarrow B=\left(\frac{1}{25}-\frac{1}{75}\right):2\)
\(\Rightarrow B=\frac{1}{75}\)
Vậy \(B=\frac{1}{75}\)
\(C=\frac{4}{2.4}+\frac{4}{4.6}+...+\frac{4}{2008.2010}\)
\(\Rightarrow\frac{2}{4}C=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\)
\(\Rightarrow\frac{2}{4}C=\frac{1}{2}-\frac{1}{2010}=\frac{502}{1005}\)
\(\Rightarrow C=\frac{502}{1005}:\frac{2}{4}=\frac{1004}{1005}\)
Vậy \(C=\frac{1004}{1005}\)
Ủng hộ tớ nha m.n ^_^
A = \(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
=\(7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{70}\right)\)
=\(7.\frac{3}{35}\)
=\(\frac{3}{5}\)
B=\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
=\(\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)
=\(\frac{1}{2}.\frac{2}{75}\)
=\(\frac{1}{75}\)
a,
suy ra A = 7. (1/10.11+1/11.12+1/12.13+.......+1/69.70)
suy ra A = 7. ( 1/10 - 1/11+ 1/11 - 1/12 + 1/12 - 1/13+ ............. + 1/69 - 1/70)
suy ra A = 7. ( 1/ 10 - 1/70)
suy ra A= 7. 3/35
suy ra A= 3/5
\(A=\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{100\cdot104}\)
\(A=\frac{7-4}{4\cdot7}+\frac{11-7}{7\cdot11}+\frac{15-11}{11\cdot15}+...+\frac{104-100}{100\cdot104}\)
\(A=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)
\(A=\frac{1}{4}-\frac{1}{104}\)
\(A=\frac{25}{104}\)
\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)
\(B\cdot2=\left(\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\right)\cdot2\)
\(B\cdot2=\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\)
\(B\cdot2=\frac{27-25}{25\cdot27}+\frac{29-27}{27\cdot29}+\frac{31-29}{29\cdot31}+...+\frac{75-73}{73\cdot75}\)
\(B\cdot2=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\)
\(B\cdot2=\frac{1}{25}-\frac{1}{75}\)
\(B\cdot2=\frac{2}{75}\)
\(B=\frac{2}{75}\frac{\cdot}{\cdot}2\)
\(B=\frac{1}{75}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(2A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(2A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(2A=\frac{1}{1}-\frac{1}{100}\)
\(2A=\frac{99}{100}\Rightarrow A=\frac{99}{100}:2\Rightarrow A=\frac{99}{200}\)
Câu B và C làm tương tự.
bạn Nhi làm sai rồi
\(\frac{2}{2\cdot3}\) sao có thể bằng \(\frac{1}{2}-\frac{1}{3}\) được
\(\frac{1}{2\cdot3}\) mới bằng \(\frac{1}{2}-\frac{1}{3}\)
kết quả là : \(\frac{49}{100}\)
\(A=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=\dfrac{7.60}{700}=\dfrac{420}{700}=\dfrac{3}{5}\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{75}\right)=\dfrac{1}{75}\)
Ta có :
\(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
\(2B=\frac{2}{25.27}+\frac{1}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\)
\(2B=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+....+\frac{1}{73}-\frac{1}{75}\)
\(2B=\frac{1}{25}-\frac{1}{75}\)
\(2B=\frac{2}{75}\)
\(\Rightarrow B=\frac{1}{75}\)
Vậy B = \(\frac{1}{75}\)
\(F=\frac{4}{2.3}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(\Rightarrow F=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)
\(\Rightarrow F=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(\Rightarrow F=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(\Rightarrow F=2.\frac{502}{1005}=\frac{1004}{1005}\)
Vậy F = \(\frac{1004}{1005}\)