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16 tháng 7 2017

\(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{x\left(x+2\right)}=\dfrac{2011}{2013}\)

\(\Leftrightarrow2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{2011}{2013}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x-2}=\dfrac{2011}{4026}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x-2}=\dfrac{2011}{4026}\)\(\Leftrightarrow\dfrac{1}{x-2}=\dfrac{1}{2013}\)

\(\Rightarrow x-2=2013\Rightarrow x=2015\)

\(a,\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(\frac{11}{15}x=\frac{2}{5}\)

\(x=\frac{6}{11}\)

b,\(\left(2x-3\right).\left(6-2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\6-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)

Vậy

17 tháng 4 2019

2

\(S1=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)

\(S1=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\right)\)

\(S1=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\right)\)

\(S1=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{102}\right)\)

\(S1=\frac{1}{2}.\left(\frac{51}{102}-\frac{1}{102}\right)\)

\(S1=\frac{1}{2}.\frac{25}{51}\)

\(S1=\frac{25}{102}\)

Sửa đề: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)

Ta có: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)

\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2018\cdot2020}+\dfrac{2}{2020\cdot2022}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2018}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)

\(=2\cdot\dfrac{505}{1011}\)

\(=\dfrac{1010}{1011}\)

9 tháng 5 2018

\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x.\left(x+2\right)}=\dfrac{4}{9}\)

=\(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x-2}\)

=\(\dfrac{1}{2}-\dfrac{1}{x+2}\)=\(\dfrac{4}{9}\)

=>\(\dfrac{1}{x+2}=\dfrac{1}{2}-\dfrac{4}{9}=\dfrac{1}{18}\)

=>\(x+2=18\)

=>x=16

16 tháng 4 2016

K=2.(2/2.4+2/4.6+2/6.8+...+2/2008.2010)

K=2.(4-2/2.4+6-4/4.6+8-6/6.8+...+2010-2008/2008.2010)

K=2.(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010)

K=2.(1.2-1.2010)

K=2.502/1005

K=1004/1005

1 tháng 4 2015

F=4/2.4+4/4.6+4/6.8+..........+4/2008.2010

F=2/2-2/4+2/4-2/6+2/6-2/8+......+2/2008-2/2010

F=2/2- 2/4+2/4-2/6+2/6-2/8+......+2/2008-2/2010

F=2/2-2/2010

=>F=2008/2010=1004/1005

1 tháng 4 2015

\(\frac{1004}{1005}\)

4 tháng 4 2017

Gọi F= 4/2.4+4/4.6+4/6.8+...+4/2008.2010
F/2= 2/2.4+2/4.6+...+2/2008.2010
Mà 2/2.4=1/2-1/4; 2/4.6=1/4-1/6 ....
Vậy F/2= (1/2-1/4)+(1/4-1/6)+....+(1/2008-1/2010)
F/2=1/2-1/2010=2010/4020-2/4020=2008/4...
F= 2008.2/4020=1004/1005