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\(\left(-\frac{1}{2}\right)+\left(-\frac{1}{9}\right)-\left(-\frac{3}{5}\right)+\frac{1}{2006}-\left(-\frac{2}{7}\right)\)
\(=\left(-\frac{1}{2}\right)-\frac{1}{9}+\frac{3}{5}+\frac{1}{2006}+\frac{2}{7}\)
\(=\left[\left(-\frac{1003}{2006}+\frac{1}{2006}\right)\right]+\left[\left(-\frac{5}{45}+\frac{27}{45}\right)\right]+\frac{2}{7}\)
\(=-\frac{1002}{2006}+\frac{22}{45}+\frac{2}{7}\)
\(=-\frac{501}{1003}+\frac{154}{315}+\frac{90}{315}\)
\(=-\frac{501}{1003}+\frac{244}{315}\)
\(=-\frac{157815}{315945}+\frac{244732}{315945}=\frac{86917}{315945}\approx0,28\)
P/s : Vì bài này số quá xấu nên mình đổi ra số thập phân cho gọn nhé ....
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) =
\(C=\frac{\left(1+\frac{1999}{1}\right)\left(1+\frac{1999}{2}\right)...\left(1+\frac{1999}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)...\left(1+\frac{1000}{1999}\right)}\)=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{1001.1002.1003....2999}{1.2.3...1999}}\)
=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}}\)
=> \(C=\frac{2000.2001.2002....2999}{1.2.3...1000}.\frac{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}=1\)
Đáp số: C=1
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+100}\right)\)
\(A=\frac{2}{\left(1+2\right).2:2}.\frac{5}{\left(1+3\right).3:2}.\frac{9}{\left(1+4\right).4:2}...\frac{5049}{\left(1+100\right).100:2}\)
\(A=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{10098}{100.101}\)
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{99.102}{100.101}\)
\(A=\frac{1.2.3...99}{2.3.4...100}.\frac{4.5.6...102}{3.4.5...101}\)
\(A=\frac{1}{100}.\frac{102}{3}=100.34=\frac{1}{100}.34=\frac{17}{50}\)
1, A=\(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{100}{99}\)
A= \(\frac{100}{2}\)
A=50
2, B=\(\frac{-1}{2}.\frac{-2}{3}....\frac{-98}{99}\)
B= \(\frac{1}{99}\)
\(A=\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)......\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}......\frac{99}{98}\cdot\frac{100}{99}\)
\(=\frac{100}{2}\)
\(=50\)
\(B=\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)......\left(\frac{1}{99}-1\right)\)
\(=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot\left(-\frac{3}{4}\right).....\left(-\frac{97}{98}\right)\cdot\left(-\frac{98}{99}\right)\)
\(=-\frac{1}{99}\)
\(A=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)......\left(1-\frac{1}{1+2+3+...+2006}\right)\)
\(A=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)....\left(1-\frac{1}{2013021}\right)\)
\(A=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}....\frac{2013020}{2013021}\)
\(A=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}......\frac{4026040}{4026042}\)
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}......\frac{2005.2008}{2006.2007}\)
\(A=\frac{1.2.3.....2005}{2.3.4....2006}.\frac{4.5.6....2008}{3.4.5...2007}\)
\(A=\frac{1}{2006}.\frac{2008}{3}=\frac{1004}{3009}\)