2/2.3+2/3.4+2/4.+.......+2/x.(x+1)=2013/2015
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16 tháng 8 2016
\(\frac{2}{1.2}+\frac{2}{2.3}+..........+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}\)
\(\Rightarrow2\left(\frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{x\left(x+1\right)}\right)=\frac{4028}{2015}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..........+\frac{1}{x}-\frac{1}{x+1}=\frac{4028}{2015}:2\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2014}{2015}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2014}{2015}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2015}\)
\(\Rightarrow x+1=2015\Rightarrow x=2014\)
OP
16 tháng 8 2016
\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}\)
\(2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x\times\left(x+1\right)}\right)=1\frac{2013}{2015}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=1\frac{2013}{2015}\div2\)
\(1-\frac{1}{x+1}=\frac{2014}{2015}\)
\(\frac{1}{x+1}=1-\frac{2014}{2015}\)
\(\frac{1}{x+1}=\frac{1}{2015}\)
\(x+1=2015\)
\(x=2015-1\)
\(x=2014\)
28 tháng 12 2015
2/1.2+2/2.3+2/3.4+...+2/x(x+1)=4028/2015
2(1/1.2+1/2.3+1/3.4+...+1/x(x+1))=4028/2015
2(1/1-1/2+1/2-1/3+1/3-1/4+....+1/x-1/x+1)=4028/2015
2(1-1/x+1)=4028/2015
1-1/x+1=2014/2015
(x+1-1)/x+1=2014/2015
x/x+1=2014/2015
(x+1).2014=2015x
2014x-2015x=-2014
-x=-2014
x=2014
16 tháng 3 2017
gọi A là tên biểu thức vế trái
Ta có : A = 2/2.3 + 2/3.4 + 2/4.5 + ... + 2/x(x+1)
A = 2 . 1/2.3 + 2 . 1/3.4 + 2 . 1/4.5 + ... + 2 . 1/x(x+1)
A = 2 . ( 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/x(x+1)
A = 2 . ( 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1 )
A = 2 . ( 1/2 - 1/x+1 )
thay vào được :
2 . ( 1/2 - 1/x+1 ) = 2015/2017
1/2 - 1/x+1 = 2015/4034
1/x+1 = 1/2017
=> x + 1 = 2017
=> x = 2016
HN
0
\(\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2013}{2015}\)
=>\(\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+...+\dfrac{2}{x}-\dfrac{2}{x+1}=\dfrac{2013}{2015}\)
=>\(1-\dfrac{2}{x+1}=\dfrac{2013}{2015}\)
=>\(\dfrac{2}{x+1}=\dfrac{2}{2015}\)
=>x+1=2015
=>x=2014