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ta có tử số bằng :{2008 +2007/2 +... 2+1/2008} = {2007/2 +1 +2006/3+1 +...+1/2008+1} = {2009/2 +2009/3 +...+2009/2008} =
2009x{1/2 +1/3 +1/4+...+1/2009} . Vậy A = 2009
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giải dùm mk với. chiều ni mk thi học kì. bài cuối sợ là bài này lắm
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câu hỏi hay......nhưng tui xin nhường cho các bn khác
Hãy tích đúng cho tui nha
THANKS
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\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)
\(=\frac{1}{5}+\frac{2}{3}\)
\(=\frac{13}{15}\)
\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{2009}-1\right)\\ =\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\left(\dfrac{1}{3}-\dfrac{3}{3}\right)...\left(\dfrac{1}{2009}-\dfrac{2009}{2009}\right)\\ =\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\dfrac{-2008}{2009}\\ =\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2008}{2009}\\ =\dfrac{2\cdot3\cdot...\cdot2008}{\left(2\cdot3\cdot...\cdot2008\right)\cdot2009}\\ =\dfrac{1}{2009}\)