1/1×2+1/2×3+1/3×4+.....1/2006×2007
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\(\frac{M}{N}=\frac{\frac{1}{2007}+\frac{2}{2006}+......+\frac{2006}{2}+\frac{2007}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{2006}+\frac{1}{2007}}\)
\(\frac{M}{N}=\frac{\frac{1}{2007}+1+\frac{2}{2006}+1+.......+\frac{2007}{1}+1+\frac{2008}{2008}-2008}{\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+.....+\frac{1}{2}}\)
\(\frac{M}{N}=\frac{\frac{2008}{2007}+\frac{2008}{2006}+....+\frac{2008}{1}+\frac{2008}{2008}-2008}{\frac{1}{2008}+........+\frac{1}{2}}\)
đến đây là ra rùi ha
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot......\cdot\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot......\cdot\frac{19}{20}\)
\(A=\frac{1.2.3.....19}{2.3........20}\)
\(A=\frac{1}{20}\)
Xét tử
2008+2007/2+2006/3+2005/4+ ... +2/2007+1/2008
=(1+1+1+...+1)+2007/2+2006/3+2005/4+ ... +2/2007+1/2008
= 1+ (2007/2)+1+(2006/3)+1+(2005/4)+1+ ... + (2/2007)+1+(1/2008)+1
=2009/2009+2009/2+2009/3+2009/4+ ... + 2009/2007 + 2009/2008
=2009.(1/2+1/3+1/4+ ... + 1/2007+1/2008+1/2009)
=1/1-1/2+1/2-1/3+1/3-1/4+..........+1/2006-1/2007
=1/1-1/2007
=2006/2007
OKnha bn
\(\frac{1}{1x2}\)\(+\)\(\frac{1}{2x3}\)\(+\)\(\frac{1}{3x4}\)\(+\)\(...\)\(+\)\(\frac{1}{2006x2007}\)
\(=\) \(\frac{2-1}{1x2}\)\(+\)\(\text{}\)\(\frac{3-2}{2x3}\)\(+\)\(\frac{4-3}{3x4}\)\(+\) \(....\)\(+\)\(\frac{2007-2006}{2006x2007}\).
\(=\) \(1\) \(-\)\(\frac{1}{2}\)\(+\) \(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)\(....\)\(+\)\(\frac{1}{2006}\)\(-\)\(\frac{1}{2007}\)
\(=\)\(1-\)\(\frac{1}{2007}\)
\(=\frac{2006}{2007}\)\(.\)
Mong đc k !! HT~
Nhưng bài này mk lớp 5 cũng lm đc sao lại nói là toán lớp 7 ?!