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Ta có:A=(1−12)×(1−13)×.......×(1−12000)A=(1−12)×(1−13)×.......×(1−12000)
=12×23×34×..........×19992000=12×23×34×..........×19992000
=12000=12000
Vậy giá trị của A là:12000
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2001}{2000}=\dfrac{2001}{2}\)
a: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2001}{2000}=\dfrac{2001}{2}\)
b: \(=101\left(34+13-27\right)=101\cdot20=2020\)
c: \(=24\%+8\%+59\%+9\%=1\)
\(\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}+\frac{1}{4}\times\frac{1}{5}+\frac{1}{5}\times\frac{1}{6}\)
\(=\frac{1}{2\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{2}-\frac{1}{6}\)
\(=\frac{1}{3}\)
M = \(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{99x100}\)
M = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
M = \(1-\dfrac{1}{100}\)
M = \(\dfrac{99}{100}\)
\(M=1\times\dfrac{1}{2}+\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+...+\dfrac{1}{99}\times\dfrac{1}{100}\)
\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(M=1-\dfrac{1}{100}\)
\(M=\dfrac{99}{100}\)
1) =1/2 x 2/3 x 3/4 x 4/5 x .... x 2002/2003 x 2003/2004
=1/2004
2) 1/2 x X-3/4=5/6
1/2 x X =3/4+5/6
1/2 x X =19/12
X=19/6
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2002}{2003}.\frac{2003}{2004}\)
\(=\frac{1.2.3...2002.2003}{2.3.4...2003.2004}=\frac{1}{2004}\)
\(\frac{1}{2}.x-\frac{3}{4}=\frac{5}{6}\)
\(\frac{1}{2}.x=\frac{5}{6}+\frac{3}{4}\)
\(\frac{1}{2}.x=\frac{10}{12}+\frac{9}{12}=\frac{19}{12}\)
\(x=\frac{19}{12}:\frac{1}{2}\)
\(x=\frac{19}{12}.2=\frac{19}{6}\)
Bài 3:
= 1- 1/2 + 1/2 -1/3 +...+ 1/98 -1/99
= 1- 1/99
= 98/99
Bài 4:
= 1/2*3 + 1/3*4 + 1/4*5 +...+ 1/10*11
= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +...+ 1/10 - 1/11
= 1/2 - 1/11= 9/22
1000,5
\(\left(\dfrac{2+1}{2}\right)\left(\dfrac{3+1}{3}\right)\left(\dfrac{4+1}{4}\right)...\left(\dfrac{2000+1}{2000}\right)\)
\(=\dfrac{3.4.5...2001}{2.3.4...2000}=\dfrac{2001}{2}\)