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Đặt biểu thức là A ta có:
\(A=\frac{\frac{2006}{2}+\frac{2006}{3}+\frac{2006}{4}+...+\frac{2006}{2007}}{\frac{2006}{1}+\frac{2005}{2}+\frac{2004}{3}+...+\frac{1}{2006}}\)
\(\Rightarrow A=\frac{2006.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}\right)}{1+\left(1+\frac{2005}{2}\right)+\left(1+\frac{2004}{3}\right)+...+\left(1+\frac{1}{2006}\right)}\)
\(\Rightarrow A=\frac{2006.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}\right)}{1+\frac{2007}{2}+\frac{2007}{3}+...+\frac{2007}{2006}}\)
\(\Rightarrow A=\frac{2006.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}\right)}{2007.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2006}+\frac{1}{2007}\right)}\)
\(\Rightarrow A=\frac{2006}{2007}\)
A = 3 + 6 + 9 + ... + 2007
=>A = 3( 1 + 2 + 3 + ... + 669 )
=> A = \(3\cdot\left(\frac{670\cdot669}{2}\right)\)
=> A = \(3\cdot224115\)= 672345
B = \(2\cdot53\cdot12+4\cdot6\cdot87-3\cdot8\cdot40\)
=> B = 24 * 53 + 24 * 87 - 24 * 40
=> B = 24 * ( 53 + 87 - 40 )
=> B = 24 * 100 = 2400
c) ta có Tử số = \(2006\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}\right)\)
Mẫu số = \(\frac{2007-1}{1}\)+\(\frac{2007-2}{2}\)+...+\(\frac{2007-2006}{2006}\)
=> Mẫu số = \(\frac{2007}{1}\)\(-1\)+ \(\frac{2007}{2}\)\(-1\)+ ... + \(\frac{2007}{2006}\)\(-1\)
=> Mẫu số = \(\frac{2007}{1}\)+ \(\frac{2007}{2}\)+ ... + \(\frac{2007}{2006}\)- ( 1 + 1 + 1 + ... + 1 ) ( 1 + 1 + ... + 1 có 2006 số hạng 1 )
=> Mẫu số = ( 2007 - 2006 ) + \(2007\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2006}\right)\)
=> Mẫu số = \(\frac{2007}{2007}\)+ \(2007\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2006}\right)\)
=> Mẫu số = \(2007\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}\right)\)
=> C = \(\frac{TS}{MS}\)= \(\frac{2006}{2007}\)
=1+1/2001+1+1/2002+1+1/2003+...+1+1/2008=8+1/2001+1/2002+1/2003+...+1/2008>8
\(\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)
\(C=\frac{\frac{2006}{2}+\frac{2006}{3}+\frac{2006}{4}+....+\frac{2006}{2007}}{\frac{2006}{1}+\frac{2005}{2}+\frac{2004}{3}+.....+\frac{1}{2006}}\)
Đặt N = \(\frac{2006}{1}+\frac{2005}{2}+\frac{2004}{3}+.....+\frac{1}{2006}\)
\(\Rightarrow N=\frac{1}{2006}+.....+\frac{2004}{3}+\frac{2005}{2}+\frac{2006}{1}\)
\(\Rightarrow N=\left(\frac{1}{2006}+1\right)+.....+\left(\frac{2004}{3}+1\right)+\left(\frac{2005}{2}+1\right)+1\)( Có 2005 nhóm )
\(=\frac{2007}{2006}+....+\frac{2007}{3}+\frac{2007}{2}+\frac{2007}{2007}\)
\(=2007\left(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2006}+\frac{1}{2007}\right)\)
Đặt M = \(\frac{2006}{2}+\frac{2006}{3}+\frac{2006}{4}+....+\frac{2006}{2007}\)
\(=2006\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2007}\right)\)
Thay N và M vào C , ta có :
\(C=\frac{N}{M}=\frac{2006\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2007}\right)}{2007\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2007}\right)}=\frac{2006}{2007}\)
\(\Rightarrow C=\frac{2006}{2007}\)
Vậy : \(C=\frac{2006}{2007}\)