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1/
A= 1/15+1/35+1/63+1/99+ ... + 1/9999
A=1/3.5+1/5.7+1/7.9+ ... +1/99.101
2A=2/3.5+2/5.7+2/7.9+ ... +2/99.101
2A=1/3-1/5+1/5-1/7+1/7-1/9+ ... + 1/99-1/101
2A=1/3-1/101
A=49/303
Sai thì thôi nhé
A= 1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
A=1-1/7
A=6/7
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B = \(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{15}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{63}\)
B = \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{12}+\frac{1}{15}+\frac{1}{20}\right)+\left(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)+\frac{1}{63}\)
B = \(1+\frac{1}{5}+\frac{3}{40}+\frac{1}{63}\)
B = \(1\frac{11}{40}+\frac{1}{63}\)
B = \(1\frac{733}{2520}\)
nguyentuantai làm hòa với Nguyễn Đình Dũng phải chăng mục đích là lấy **** ko
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Bạn chuyển thành dạng các phân số có tử số bằng 3 bằng cách nhân mỗi phân số với 3 rồi cả tổng tất cả nhân với 1/3. Sau đó làm như bình thường
Ta có:
A\(=\frac{1}{2x5}+\frac{1}{5x3}+\frac{1}{3x7}+\frac{1}{7x4}+...+\frac{1}{14x29}+\frac{1}{29x15}\)
\(=\frac{2}{2x\left(2x5\right)}+\frac{2}{\left(5x3\right)x2}+\frac{2}{2x\left(3x7\right)}+\frac{2}{\left(7x4\right)x2}+...+\frac{2}{2x\left(14x29\right)}+\frac{2}{\left(29x15\right)x2}\)
\(=\frac{2}{4x5}+\frac{2}{5x6}+\frac{2}{6x7}+\frac{2}{7x8}+...+\frac{2}{28x29}+\frac{2}{29x30}\)
\(=2x\left(\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{7x8}+...+\frac{1}{28x29}+\frac{1}{29x30}\right)\)
\(=2x\left(\frac{5-4}{4x5}+\frac{6-5}{5x6}+\frac{7-6}{6x7}+\frac{8-7}{7x8}+...+\frac{29-28}{28x29}+\frac{30-29}{29x30}\right)\)
\(=2x\left(\frac{5}{4x5}-\frac{4}{4x5}+\frac{6}{5x6}-\frac{5}{5x6}+\frac{7}{6x7}-\frac{6}{6x7}+\frac{8}{7x8}-\frac{7}{7x8}+...+\frac{29}{28x29}-\frac{28}{28x29}+\frac{30}{29x30}-\frac{29}{29x30}\right)\)
\(=2x\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{28}-\frac{1}{29}+\frac{1}{29}-\frac{1}{30}\right)\)
\(=2x\left(\frac{1}{4}-\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{6}\right)-...-\left(\frac{1}{28}-\frac{1}{28}\right)-\left(\frac{1}{29}-\frac{1}{29}\right)-\frac{1}{30}\right)\)
\(=2x\left(\frac{1}{4}-0-0-...-0-0-\frac{1}{30}\right)\)
\(=2x\left(\frac{1}{4}-\frac{1}{30}\right)\)
\(=2x\frac{1}{4}-2x\frac{1}{30}\)
\(=\frac{1}{2}-\frac{1}{15}\)
=15/30-2/30=13/30
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Xét tử ta có:
\(2008+\frac{2007}{2}+\frac{2006}{3}+....+\frac{1}{2008}\)
= \(1+\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{1}{2008}\right)\)
= \(\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2008}\)
= \(2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)\)
=> A = \(\frac{2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}}\)
=> A = 2009
A=\(\frac{\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+\left(1+\frac{2005}{4}\right)+...........+\left(1+\frac{2}{2008}\right)+\left(1+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2008}+\frac{1}{2009}}\)=\(\frac{\frac{2009}{2}+\frac{2009}{3}+\frac{2009}{4}+....+\frac{2009}{2008}+\frac{2009}{2009}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\frac{ }{ }\)
=\(\frac{2009\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2008}+\frac{1}{2009}}\frac{ }{ }\)
=2009
Vay A=2009