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uk . mk thấy bạn đăng nhưng ko ai trả lời thì mk đăng hộ vs cả bài này mk cũng biết làm hihi
Đặt: \(L_2=\dfrac{2007}{1}+\dfrac{2006}{2}+\dfrac{2005}{3}+...+\dfrac{2}{2006}+\dfrac{1}{2007}\)
\(L_2=1+\left(\dfrac{2006}{2}+1\right)+\left(\dfrac{2005}{3}+1\right)+...+\left(\dfrac{2}{2006}+1\right)+\left(\dfrac{1}{2007}+1\right)\)
\(L_2=\dfrac{2008}{2008}+\dfrac{2008}{2}+\dfrac{2008}{3}+...+\dfrac{2008}{2006}+\dfrac{2008}{2007}\)
\(L_2=2008\left(\dfrac{1}{2}+\dfrac{1}{3}+..+\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)
\(\dfrac{L_1}{L_2}=\dfrac{1}{2008}\)
số số hạng của A là :
( 2007 - 3 ) : 3 + 1 = 669 ( số )
tổng A là :
( 2007 + 3 ) . 669 : 2 = 672345
B = \(\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)
B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\left(\dfrac{2005}{2}+1\right)+\left(\dfrac{2004}{3}+1\right)+...+\left(\dfrac{1}{2006}+1\right)+1}\)
B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2006}+\dfrac{2007}{2007}}\)
B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{2007.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2006}+\dfrac{1}{2007}\right)}\)
B = \(\dfrac{2006}{2007}\)
2006/1 là 2006, tách 1 của 2006 ra 2005 phân số còn lại 1
a) \(21^{10}-1=\left(21^5\right)^2-1^2=\left(21^5+1\right).\left(21^5-1\right)\)
\(21^5+1=\overline{...1}=2k+1+1=2n\)
\(21^5-1=\overline{...01}-1=\overline{...00}\)
\(\Rightarrow21^{10}-1=2n.\overline{...00}⋮200\left(đpcm\right).\)
b) \(39\equiv-1\left(mod40\right)\)
\(\Rightarrow39^{20}\equiv1\left(mod40\right)\)
\(\Rightarrow39^{19}\equiv-1\left(mod40\right)\)
\(\Rightarrow39^{20}+39^{19}\equiv1+\left(-1\right)\left(mod40\right)\)
\(\Leftrightarrow39^{20}+39^{19}\equiv0\left(mod40\right)\)
\(\Rightarrow39^{20}+39^{19}⋮40\left(đpcm\right).\)
d) \(2005\equiv-1\left(mod2006\right)\)
\(\Rightarrow2005^{2007}\equiv\left(-1\right)^{2007}=-1\left(mod2006\right)\)
\(2007\equiv1\left(mod2006\right)\)
\(\Rightarrow2007^{2005}\equiv1\left(mod2006\right)\)
\(\Rightarrow2005^{2007}+2007^{2005}\equiv-1+1=0\left(mod2006\right)\)
\(\Leftrightarrow2005^{2007}+2007^{2005}⋮2006\left(đpcm\right).\)