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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4+5}\) = \(\frac{1}{3}+\frac{1}{6}+\frac{1}{15}\)= \(\frac{10}{30}+\frac{5}{30}+\frac{2}{30}\)= \(\frac{10+5+2}{30}\)= \(\frac{17}{30}\)
Trong đề có 4 đáp án là a)\(\frac{1}{6}\)
b)\(\frac{5}{6}\)
c) \(\frac{1}{3}\)
d)\(\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2008 + 2007/2 + 2006/3 + 2005/4 + ... + 2/2007 + 1/2008
2009-1/1 + 2009-2/2 + 2009-3/3 + 2009-4/4 + ... + 2009-2007/2007 + 2009-2008/2008
2009 - 1 + 2009/2 - 1 + 2009/3 - 1 + 2009/4 - 1 + ... + 2009/2007 - 1 + 2009/2008 - 1
2009 + 2009.(1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 ) - ( 1 + 1 + 1 + 1 + ... + 1 + 1 )
2009 + 2009.( 1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 ) - 2008
1 + 2009.( 1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 )
2009.( 1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 + 1/2009 )
=> giá trị của biểu thức trên là 2009
![](https://rs.olm.vn/images/avt/0.png?1311)
1/1+2+1/1+2+3+1/1+2+3+4+1/1+2+3+4+5
=1/3+1/6+1/10+1/15
lấy mẫu số chung là 30,ta có:
10/30+5/6+3/10+2/30
=20/30=2/3
![](https://rs.olm.vn/images/avt/0.png?1311)
2008=1+1+1+...+1 có 2008 số 1
1+(1+2007/2)+(1+2006/3)+...+(1+1/2008)=2009/2009+2009/2+2009/3+...+2009/2008
=2009*(1/2009+1/2+1/3+...+1/2008)=2009*(1/2+1/3+...+1/2009)
ta có 2008+2007/2+...+1/2008
1/2+1/3+..............+1/2009
=2009
![](https://rs.olm.vn/images/avt/0.png?1311)
a,11/5x3/2-3/1/2x3/5
=11/2x3/5-7/2-3/5
=3/5x(11/2-7/2)
=3/5x2
=6/5
b,(1/2-1/3+1/4-1/5):(1/4-1/6)
=(1/6+1/20):1/12
=13/60:1/2
=13/30
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{2008+\left(1+\frac{2007}{2}\right)+...+\left(1+\frac{1}{2008}\right)}{\frac{1}{2}+...\frac{1}{2009}}-2007\)
\(=\frac{1+\frac{2009}{2}+...\frac{2009}{2008}}{\frac{1}{2}+...+\frac{1}{2009}}\)
\(=\frac{\frac{2009}{2009}+\frac{2009}{2}+...+\frac{2009}{2008}}{\frac{1}{2}+...+\frac{1}{2009}}\)
\(=\frac{2009\left(\frac{1}{2}+...+\frac{1}{2009}\right)}{\frac{1}{2}+...+\frac{1}{2009}}=2009\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)
\(M=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}\)
\(M=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+\frac{1}{\left(1+5\right).5:2}\)
\(M=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}\)
\(M=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\right)\)
\(M=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)
\(M=2.\left(\frac{1}{2}-\frac{1}{6}\right)\)
\(M=2.\frac{1}{2}-2.\frac{1}{6}\)
\(M=1-\frac{1}{3}=\frac{2}{3}\)