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\(P=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.......+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(P=2.\left(1-\frac{1}{2014}\right)\)
\(P=2.\frac{2013}{2014}\)
\(P=\frac{2013}{1007}\)
\(P=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2013.2014}\)
\(P=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)
\(P=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(P=\frac{1}{2}\left(1-\frac{1}{2014}\right)\)
\(P=\frac{1}{2}.\frac{2013}{2014}\)
\(P=\frac{2013}{4028}\)
\(=\frac{1.2}{99.100}\)
\(=\frac{2}{9900}=\frac{1}{4950}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2014.2015.2016}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2015.2016}\right)\)
= 1/2 .( 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + .......+ 1/2014.2015 - 1/2015.2016)
= 1/2 ( 1/2 - 1/2015.2016)
Tính tiếp p nhé.
\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{9}-\frac{1}{10}\)
\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{10}=\frac{9}{10}\)
\(\frac{2}{5}x=\frac{9}{10}-\frac{3}{10}=\frac{3}{5}\)
\(x=\frac{\frac{3}{5}}{\frac{2}{5}}=\frac{3}{2}\)
Ta có: \(\frac{1}{1x2}\)+ \(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ \(\frac{1}{4x5}\)+ .....+ \(\frac{1}{9x10}\)
= \(1-\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
= 1 - \(\frac{1}{10}\)
= \(\frac{9}{10}\)
\(3C=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{27.28.29.30}\)
\(3C=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+...+\frac{30-27}{27.28.29.30}\)
\(3C=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}+\frac{1}{28.29.30}\)
\(3C=\frac{1}{1.2.3}-\frac{1}{28.29.30}\Rightarrow C=\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right):3\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
= 1 / 1 - 1 / 2 + 1 / 2 - 1 / 3 + 1 / 3 - 1 / 4 + 1 / 4 - 1 / 5 + 1 / 5 - 1 / 6
Ta gạch các ps trùng.
Còn lại :
1 / 1 - 1 / 6 = 6 / 5
ĐẶT BIỂU THỨC LÀ A
Ta có công thức : \(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)
Dựa vào công thức, ta có :
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(A=2.\left(1-\frac{1}{2014}\right)=2.\frac{2013}{2014}=\frac{2013}{1007}\)
Ai thấy đúng thì ủng hộ nha !!!
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2013.2014}=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)
\(=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(=2.\left(\frac{1}{1}-\frac{1}{2015}\right)=2.\left(\frac{2015}{2015}-\frac{1}{2015}\right)=2.\frac{2014}{2015}=\frac{4028}{2015}\)