Tìm x biết \(\frac{4}{2\times4}+\frac{4}{4\times6}+\frac{4}{6\times8}+...+\frac{4}{2008\times2010}+x=\frac{-1}{1005}\)
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\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
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\(=\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+...+\frac{2}{12\times14}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{12}-\frac{1}{14}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{2}\times\frac{3}{7}=\frac{3}{14}\)
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\(S.2=\frac{2}{2.4}+\frac{2}{4.6}+...........+\frac{2}{98.100}\)
\(S.2=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{98}-\frac{1}{100}\)
\(S.2=\frac{1}{2}-\frac{1}{100}\)
\(S.2=\frac{49}{100}\)
\(S=\frac{49}{100}:2\)
\(S=\frac{49}{200}\)
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:V Làm sai hết rồi sai ngay từ bước đầu tiên.
\(\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-....-\frac{1}{9.10}\)
\(=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}\right)\)
\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(=\frac{1}{12}-\frac{3}{20}\)
\(=\frac{-11}{12}\)
\(\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{9.10}\)
= \(-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
= \(-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
= \(-\left(\frac{1}{3}-\frac{1}{10}\right)\)
= \(-\frac{7}{30}\)
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1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6+1/7-1/7+1/8-1/8+1/9+1/9-1/10
=1/2-1/10
=5/10-1/10
=4/10=2/5
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{8x9}+\frac{1}{9x10}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{2}-\frac{1}{10}\)
\(\frac{2}{5}\)
\(\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+...+\frac{4}{2008\cdot2010}+x=-\frac{1}{1005}\)
\(2\cdot\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{2008\cdot2010}\right)+x=-\frac{1}{1005}\)
\(2\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)+x=-\frac{1}{1005}\)
\(2\cdot\left[\left(\frac{1}{2}-\frac{1}{2010}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{6}-\frac{1}{6}\right)+...+\left(\frac{1}{2008}-\frac{1}{2008}\right)\right]+x=-\frac{1}{1005}\)\(2\cdot\left[\left(\frac{1005}{2010}-\frac{1}{2010}\right)+0+...+0\right]+x=-\frac{1}{1005}\)
\(2\cdot\frac{1004}{2010}+x=-\frac{1}{1005}\)
\(\frac{1004}{1005}+x=-\frac{1}{1005}\)
\(x=-\frac{1}{1005}-\frac{1004}{1005}=-1\)
Vậy x=-1
Chúc bạn học tốt!^_^