Tính:\(\frac{2}{2\cdot4}\)+\(\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2014\cdot2016}=\)
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ĐKXĐ: \(x\ne0;x\ne-2\)
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{1}{x+2}=\frac{1}{18}\)
\(\Rightarrow\)\(x+2=18\)
\(\Leftrightarrow\)\(x=16\) (t/m ĐKXĐ)
Vậy...
1/2(1-1/4+1/4-1/6+1/6-1/8+...+1/x-1/x+2)=4/9
1/2(1-1/x+2)=4/9
1- 1/x+2=4/9:1/2
1 - 1 /x+2=8/9
1/x+2=1-8/9
1/x+2=1/9
suy ra x+2=9
x=9-2
x=7
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2012.2014}\)
\(\Leftrightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2012}-\frac{1}{2014}\right)\)
\(\Leftrightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2014}\right)\)
\(\Leftrightarrow A=\frac{1}{2}\cdot\frac{503}{1007}\)
\(\Leftrightarrow A=\frac{503}{2014}\)
= 1/2[1/2 - 1/4+1/4-1/6 + 1/6-1/8+...+ 1/2012-1/2014]
= 1/2[1/2-1/2014]
= 1/2 * 503/1007
= 503/2014
\(E=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{2016.2018}\)
\(E=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{2018-2016}{2016.2018}\)
\(2E=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2016}-\frac{1}{2018}\)
\(E=\left(\frac{1}{2}-\frac{1}{2018}\right).\frac{1}{2}\)
\(E=\frac{504}{1009}.\frac{1}{2}\)
\(E=\frac{252}{1009}\)
\(E=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2016}-\frac{1}{2018}\)
\(E=\frac{1}{2}-\frac{1}{2018}\)
\(E=\frac{1005}{2018}\)
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2014.2016}=1-\frac{2}{4}+\frac{2}{4}-\frac{2}{6}+\frac{2}{6}-\frac{2}{8}+...+\frac{2}{2014}-\frac{2}{2016}\)
\(=1-\frac{2}{2016}=\frac{1007}{1008}\)