tìm x
1/3+1/6+1/10+...+2/x(x+1)=2015/2017
ai làm đúng mình xẽ tick
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DV
3
12 tháng 2 2016
(x+1)+(x+2)+.........+(x+2013)=0
=>x+1+x+2+........+x+2013=0
=>(x+x+.........+x)+(1+2+............+2013)=0
=>2013x+2013.(2013+1):2=0
=>2013x+2027091=0
=>2013x=0-2027091
=>2013x=-2027091
=>x=2027091:2013
=>x=-1007
20 tháng 4 2016
nhân cả 2 vế của đẳng thức với 1/2 ta được
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-......+\frac{1}{x}-\frac{1}{x+1}=\frac{2014}{2015}\)
\(=\frac{1}{2}-\frac{1}{x+1}=\frac{2014}{2015}\)
\(=>\frac{1}{x+1}=\frac{1}{2}-\frac{2014}{2015}\)
\(\frac{1}{x+1}=-\frac{2013}{4030}\)
hay \(1:\left(x+1\right)=-\frac{2013}{4030}\)
\(x+1=-\frac{4030}{2013}\)
\(=>x=-\frac{6043}{2013}\)
TM
11 tháng 3 2017
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{x\times\left(x+1\right)}{2}}=\frac{2015}{2017}\)
\(\Leftrightarrow\frac{1}{2}\)[\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{x\times\left(x+1\right)}{2}}\)]=\(\frac{1}{2}.\frac{2015}{2017}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\times\left(x+1\right)}=\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{2015}{4034}\)\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4034}=\frac{1}{2017}\)<=> x+1=2017<=>x=2016
CG
0
KK
10 tháng 2 2019
( x+1 )(y-2)=0
x+1=0 hoặc y-2=0
x=(-1) hoặc y=2
(x-5)(y-7)=1
x-5=1 và y-7=1
x=6 và y=8
đợi mình 1 tí mình làm tiếp
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4034}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{2}{4034}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2017}\)
\(\Leftrightarrow x+1=2017\)
\(\Leftrightarrow x=2017-1\)
\(\Leftrightarrow x=2016\)
Vậy x = 2016
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2015}{2017}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2017}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2017}\)
\(\Rightarrow2\cdot\frac{x-1}{2\left(x+1\right)}=\frac{2015}{2017}\)
\(\Rightarrow\frac{x-1}{2x+2}=\frac{2015}{4034}\)
\(\Rightarrow4034x-4034=4030x+4030\)
\(\Rightarrow4034x-4030x=8064\)
\(\Rightarrow x=2016\)