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\(B=\left(1+\dfrac{1}{100}\right)\times\left(1+\dfrac{1}{99}\right)\times....\times\left(1+\dfrac{1}{3}\right)\times\left(1+\dfrac{1}{2}\right)\)
\(B=\dfrac{101}{100}\times\dfrac{100}{99}\times...\times\dfrac{4}{3}\times\dfrac{3}{2}\)
\(B=\dfrac{101\times100\times....\times4\times3}{100\times99\times....\times3\times2}\)
\(B=\dfrac{101}{2}\)
\(\Rightarrow B=\left(\dfrac{100}{100}+\dfrac{1}{100}\right)\times\left(\dfrac{99}{99}+\dfrac{1}{99}\right)\times...\times\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\times\left(\dfrac{2}{2}+\dfrac{1}{2}\right)\)
\(B=\dfrac{101}{100}\times\dfrac{100}{99}\times...\times\dfrac{4}{3}\times\dfrac{3}{2}\)
\(B=\dfrac{101}{2}\)( triệt tiêu các mẫu, tử giống nhau)
cho 1 1/20 x 1 1/21 x ..... x 1 1/2013=21/20 x 22/21 x 23/22 x ...... x 2014/2013
=(21 x 22 x .... x 2014) / (20 x 21 x ... x 2013) =2014/20=....
\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\cdot....\cdot\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot....\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{2\cdot3\cdot4\cdot...\cdot2002\cdot2003}{3\cdot4\cdot5\cdot...\cdot2003\cdot2004}=\frac{1}{1002}\)
\(=\frac{1.2.3...2006}{2.3.4...2007}\)
Sau khi rút gọn ta được
\(=\frac{1}{2007}\)
100% đúng đó!!!
B= (1-1/2). ( 1-1/3).(1-1/4).(1-1/5)....(1-1/2004)
B= 1/2. 2/3 . 3/4. 4/5....2003/2004
B= 1/2004
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(B=\frac{1}{2004}\)
\(B=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x...x\frac{2002}{2003}x\frac{2003}{2004}\)
\(B=\frac{1x2x3x4x...x2002x2003}{2x3x4x5x...x2003x2004}\)
Rút gọn các thừa số ở tử và mẫu ta được:
\(B=\frac{1}{2004}\)
Đ/S:\(\frac{1}{2004}\)
Ta có:
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{2002}{2003}.\frac{2003}{2004}\)
\(=\frac{1.2....2002.2003}{2.3....2003.2004}\)
Đơn giản hết sẽ là:
\(=\frac{1}{2004}\)
\(\left(1+\frac{1}{100}\right).\left(1+\frac{1}{99}\right).....\left(1+\frac{1}{3}\right).\left(1+\frac{1}{2}\right)\)
\(=\frac{101}{100}.\frac{100}{99}.....\frac{4}{3}.\frac{3}{2}\)
\(=\frac{101}{2}\)
\(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\times\left(1-\dfrac{1}{6}\right)\times\dots\times\left(1-\dfrac{1}{99}\right)\times\left(1-\dfrac{1}{100}\right)\) (sửa đề)
\(=\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\times\dfrac{5}{6}\times\dots\times\dfrac{98}{99}\times\dfrac{99}{100}\)
\(=\dfrac{2\times3\times4\times5\times\dots\times98\times99}{3\times4\times5\times6\times\dots\times99\times100}\)
\(=\dfrac{2}{100}\)
\(=\dfrac{1}{50}\)