\(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+..............+\frac{2}{x\cdot\left(x+2\right)}=\frac{4}{9}\)
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DP
21 tháng 7 2017
Bài 1 :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=1-\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{10}{19}\)
\(\Leftrightarrow10.\left(2x+3\right)=19\Leftrightarrow2x+3=\frac{19}{10}\)
\(\Leftrightarrow2x=\frac{19}{10}-3\Leftrightarrow2x=-\frac{11}{10}\)
\(\Leftrightarrow x=-\frac{11}{20}=-0,55\)
Bài 2 :
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
31 tháng 8 2017
\(\left(\frac{4}{2.4}+\frac{4}{4.6}+.....+\frac{4}{4020.4022}\right)x=2010\)
\(\Leftrightarrow2x\left(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{4020.4022}\right)=2010\)
\(\Leftrightarrow2x\left(\frac{1}{2}-\frac{1}{4}+.....+\frac{1}{4020}-\frac{1}{4022}\right)=2010\)
\(\Leftrightarrow2x\left(\frac{1}{2}-\frac{1}{4022}\right)=2010\)
Tự biên tự diễn
31 tháng 8 2017
Ko chép lại đề nhé
<=> 2( 2/2.4 + 2/2.6 + 2/2.8 +...+ 2/ 4020.4022) x= 2010
<=> 2( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 +....+ 1/4020- 1/4022 )x=2010
<=> ( 1/2 - 1/4022)2x = 2010
<=> ( 2011/4022 - 1/4022 )2x = 2010
<=>( 2010/4022) .2x= 2010
<=> 2x = 2010 : 2010/4022
<=> 2x = 4022
=> x = 2011
Vậy x = 2011
13 tháng 11 2015
bài này không khó. Nhưng đánh máy để giải cho bạn thì thực sự khó
N
7 tháng 6 2017
xét \(VT=\frac{2}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+......+\frac{1}{2n.\left(2n+2\right)}\right)\) (1)
\(=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+.......+\frac{2}{2n\left(2n+2\right)}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.......+\frac{1}{2n}-\frac{1}{2n+2}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2n+2}\right)=\frac{1}{4}-\frac{1}{2\left(2n+2\right)}\)
\(=\frac{1}{4}-\frac{1}{4n+4}\)
mà theo bài ra (1) = \(\frac{502}{2009}\)
<=>\(\frac{1}{4}-\frac{1}{4n+4}=\frac{502}{2009}\)
<=>\(\frac{1}{4n+4}=\frac{1}{4}-\frac{502}{2009}\)
<=>\(\frac{1}{4n+4}=\frac{1}{8036}\)
<=> 4n+4=8036
<=> 4n=8032
<=> n=2008
7 tháng 6 2017
=) \(\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2n\left(2n+2\right)}\right)=\frac{502}{2009}\)
=) \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2n}-\frac{1}{2n+2}\right)=\frac{502}{2009}\)
=) \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2n+2}\right)=\frac{502}{2009}\)
=) \(\frac{1}{2}-\frac{1}{2n+2}=\frac{502}{2009}:\frac{1}{2}=\frac{1018}{2009}\)
=) \(\frac{1}{2n+2}=\frac{1}{2}-\frac{1018}{2009}=\frac{-27}{4018}\)
=) \(\frac{-1}{-\left(2n+2\right)}=\frac{-27}{4018}\)
=) \(\frac{-27}{27.-\left(2n+2\right)}=\frac{-27}{4018}\)
=) \(27.-\left(2n+2\right)=4018\)
=) \(-\left(2n+2\right)=4018:27=\frac{4018}{27}\)
=) \(2n+2=\frac{-4018}{27}\)
=) \(2n=\frac{-4018}{27}-2=\frac{-4072}{27}\)
=) \(n=\frac{-4072}{27}:2=\frac{-2036}{27}\)
\(\)
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