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\(C=\frac{\left(1+\frac{1999}{1}\right)\left(1+\frac{1999}{2}\right)...\left(1+\frac{1999}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)...\left(1+\frac{1000}{1999}\right)}\)=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{1001.1002.1003....2999}{1.2.3...1999}}\)
=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}}\)
=> \(C=\frac{2000.2001.2002....2999}{1.2.3...1000}.\frac{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}=1\)
Đáp số: C=1
\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)..\left(1-\frac{1}{2000^2}\right)\)
\(=\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}\cdot\frac{3.5}{4^2}\cdot\cdot\cdot\cdot\frac{1998.2000}{1999^2}\cdot\frac{1999.2001}{2000^2}\)
\(=\frac{1}{2}\cdot\frac{2001}{2000}=\frac{2001}{4000}\)
\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{1999^2}\right)\left(1-\frac{1}{2000^2}\right)\)
=\(\left(\frac{4}{4}-\frac{1}{4}\right)\left(\frac{9}{9}-\frac{1}{9}\right)...\left(\frac{3996001}{3996001}-\frac{1}{3996001}\right)\left(\frac{4000000}{4000000}-\frac{1}{4000000}\right)\)
=\(\frac{3}{4}.\frac{8}{9}....\frac{3996000}{3996001}.\frac{3999999}{4000000}\)
=\(\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{1998.2000}{1999.1999}.\frac{1999.2001}{2000.2000}\)
=\(\frac{1.3.2.4.3.6.....1998.2000.1999.2001}{2.2.3.3.4.4....1999.1999.2000.2000}=\frac{1.2001}{2.2000}=\frac{2001}{4000}\)
Ta có:
\(B=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{1999}\right).\left(1+\frac{1}{2000}\right)\)
\(=\frac{3}{2}.\frac{4}{3}...\frac{2000}{1999}.\frac{2001}{2000}=\frac{3.4....2000.2001}{2.3...1999.2000}=\frac{2001}{2}\)(Tối giản)(Chắc zậy)
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+100}\right)\)
\(A=\frac{2}{\left(1+2\right).2:2}.\frac{5}{\left(1+3\right).3:2}.\frac{9}{\left(1+4\right).4:2}...\frac{5049}{\left(1+100\right).100:2}\)
\(A=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{10098}{100.101}\)
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{99.102}{100.101}\)
\(A=\frac{1.2.3...99}{2.3.4...100}.\frac{4.5.6...102}{3.4.5...101}\)
\(A=\frac{1}{100}.\frac{102}{3}=100.34=\frac{1}{100}.34=\frac{17}{50}\)
a) \(=\frac{3}{2}.\frac{4}{3}....\frac{100}{99}=\frac{100}{2}=50\)
a) =3/2 . 4/3 . 5/4 ...100/99
=\(\frac{3.4.5...100}{2.3.4..99}\)
=\(\frac{100}{2}\)
b) =
\(=\frac{-1}{2}.\frac{-2}{3}......................\frac{-1998}{1999}.\frac{-1999}{2000}\)
\(=\frac{\left(-1\right).\left(-2\right)....................\left(-1999\right)}{1.2.3........................2000}\)
\(=\frac{-1}{2000}\)
= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{1998}{1999}.\frac{1999}{2000}=\frac{1}{2000}\)
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