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\(L=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{2-1}{2}.\frac{3-1}{3}.\frac{4-1}{4}.\frac{5-1}{5}...\frac{2003-1}{2003}.\frac{2004-1}{2004}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}....\frac{2002}{2003}.\frac{2003}{2004}=\frac{1}{2004}\)
1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x + 1) = 4007/2004
2/2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x + 1) = 4007/2004
2 × (1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x + 1)) = 4007/2004
1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x + 1 = 4007/2004 : 2
1 - 1/x + 1 = 4007/2004 × 1/2
x/x + 1 = 4007/4008
=> x = 4007
\(B=\dfrac{1}{2}x\dfrac{2}{3}x\dfrac{3}{4}x...x\dfrac{2003}{2004}\)
\(B=\dfrac{1}{2004}\)
\(B=\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times....\left(1-\dfrac{1}{2003}\right)\times\left(1-\dfrac{1}{2004}\right)\)
\(B=\dfrac{1}{2}\times\dfrac{2}{3}\times....\times\dfrac{2002}{2003}\times\dfrac{2003}{2004}\)
\(B=\dfrac{1}{2004}\)
Bạn chuyển về 1 vế sau đó trừ 1 vào mỗi phân thức ta được :
\(\left(x-2005\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}-\frac{1}{2005}\right)=0\)
Vì biểu thức bên phải khác 0 nên : \(x-2005=0\)=> \(x=2005\)
\(\frac{x-5}{2000}+\frac{x-4}{2001}+\frac{x-3}{2002}=\frac{x-2}{2003}+\frac{x-1}{2004}+\frac{x}{2005}\)
\(\Leftrightarrow\frac{x-2005}{2000}+\frac{x-2005}{2001}+\frac{x-2005}{2002}=\frac{x-2005}{2003}+\frac{x-2005}{2004}+\frac{x-2005}{2005}\)
\(\Leftrightarrow\left(x-2005\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}-\frac{1}{2005}\right)=0\)
<=> x - 2005 = 0
<=> x = 2005
Vậy ...............
\(=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2002}{2003}x\frac{2003}{2004}=\frac{1x2x3x...x2002x2003}{2x3x4x...x2003x2004}=\frac{1}{2004}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2002}{2003}.\frac{2003}{2004}\)
\(=\frac{1.2.3...2002.2003}{2.3.4...2003.2004}=\frac{1}{2004}\)
a) \(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{2002}{2003}.\dfrac{2003}{2004}\)
\(=\dfrac{1}{2004}\)
b) \(B=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}.\dfrac{9}{2}-2.\dfrac{7}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\left(\dfrac{21}{2}-\dfrac{14}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{35}{6}.\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{10}{3}\)
\(=\dfrac{3}{5}\)
Ta có
\(\hept{\begin{cases}2004\cdot2003< 2004\cdot2005\\\frac{1}{2003\cdot2004}>\frac{1}{2004\cdot2005}\end{cases}}\Rightarrow2003\cdot2004-\frac{1}{2003\cdot2004}< 2004\cdot2005-\frac{1}{2004\cdot2005}\)
Vậy 2003*2004-1/2003*2004<2004*2005-1/2004*2005.
\(=\frac{1}{2}\frac{2}{3}\frac{3}{4}\frac{4}{5}\frac{5}{6}\frac{6}{7}\frac{7}{8}\frac{8}{9}....\frac{2002}{2003}\frac{2003}{2004}\)
\(=\frac{1}{2004}\)
Sửa đề :
\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).......\left(1-\dfrac{1}{2004}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}........\dfrac{2002}{2003}.\dfrac{2003}{2004}\)
\(=\dfrac{1}{2004}\)
(1-1/2)x(1-1/3)x(1-1/4)x(1-1/5)x...x(1-1/2003)x(1-1/2004)
=1/2x2/3x3/4x4/5x...x2002/2003x2003/2004
=1/2004