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Bài 2:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{4678\times4679+4680\times31+4648}{4680\times4679-4678\times4679}\)
\(\Leftrightarrow\frac{4679\times4678+4679\times31+4679}{4679\times\left(4680-4678\right)}\)
\(=\frac{4679\times\left(4678+31+1\right)}{4679\times2}\)
\(=\frac{4679\times4710}{4679\times2}\)
\(=\frac{4710}{2}\)
\(=2355\)
\(\frac{4678\cdot4679+4680\cdot31+4648}{4680\cdot4679-4678.4679}\)=\(\frac{4678\cdot4679+\left(4680-1\right)\cdot31+4648+31}{4680\cdot4679-4678.4679}\)
=\(\frac{4678\cdot4679+4679\cdot31+4679}{4680\cdot4679-4678.4679}\)=\(\frac{\left(4678+31+1\right)\cdot4679}{\left(4680-4678\right).4679}\)=\(\frac{4710}{2}\)=2355
![](https://rs.olm.vn/images/avt/0.png?1311)
mình không biết nữa bằng bao nhiêu ấy nhỉ .......? .......? Sory ^.^
1/3 + 13/15 + 33/35 + 61/63 + 97/99
= 45/11 ( mình không tiện giải, để khi khác giải sau)
Chúc bạn may mắn!
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A = \(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}\)
\(\Rightarrow2A=2-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}\)
\(\Rightarrow2A-A=1-1+\frac{1}{16}\)
\(\Rightarrow A=\frac{1}{16}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
45*[15+1]-17/45*15+28
45*15+45-17/45*15+28
45*15+28/45*15+28
=1
k nha
![](https://rs.olm.vn/images/avt/0.png?1311)
1/1x2 + 1/2x3 +1/3x4 + ......+1/98x99+1/99x100
=1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +......+ 1/98 - 1/99 + 1/99 + 1/100
=(1-1/100)+(1/2 - 1/2 ) + ( 1/3 - 1/3 ) + ...... + (1/98 - 1/98 ) + ( 1/99 - 1/99 )
= 100/100 - 1/100 + 0 + 0 +.....+ 0 + 0
=99/100
vậy GTBT = 99/100
nhân 4 vào tử
\(\frac{1}{6.10}\)+ \(\frac{1}{10.14}\)+ ... + \(\frac{1}{402.406}\)
= \(\frac{1}{4}\). \(\left(\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{402.406}\right)\)
= \(\frac{1}{4}\). ( \(\frac{10-6}{6.10}\)+ \(\frac{14-10}{10.14}\)+ ... + \(\frac{406-402}{402.406}\))
= \(\frac{1}{4}\). ( \(\frac{10}{6.10}\)- \(\frac{6}{6.10}\)+ ... + \(\frac{406}{402.406}\)- \(\frac{402}{402.406}\))
= \(\frac{1}{4}\). ( \(\frac{1}{6}\)- \(\frac{1}{406}\))
= \(\frac{1}{4}\). \(\frac{100}{609}\)
= \(\frac{25}{609}\)