Tìm x biết: 1/3+1/6+1/10+...+1/x(x+1):2=2013/2015
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NM
0
NT
1
SN
17
2 tháng 4 2015
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\cdot\left(x+1\right)}=\frac{2013}{2015}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2013}{2015}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2013}{2015}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2013}{2015}:2\)
\(\Rightarrow-\frac{1}{x+1}=\frac{2013}{4030}-\frac{1}{2}\)
\(\Rightarrow-\frac{1}{x+1}=-\frac{1}{2015}\Rightarrow x+1=2015\Rightarrow x=2014\)
PT
1
3 tháng 4 2016
2/6+2/12+2/20+...+2/x(x+2)=2013/2015
2(1/2.3+1/3.4+...+1/x(x+1))=2013/2015
2(1/2-1/3+1/3-1/4+...+1/x-1/x+1)=2013/2015
2(1/2-1/x+1)=2013/2015
1/2-1/x+1=2013/2015:2
1/2-1/x+1=2013/4030
1/x+1=1/2-2013/4030
1/x+1=1/2015
Suy ra x+1=2015
x=2014
Vậy x=2014
5 tháng 5 2016
2/6+2/12+2/20+...+2/x.(x+1)=2013/2015
2.[1/6+1/12+1/20+...+1/x.(x+1)]=2013/2015
1/2.3+1/3.4+1/4.5+...+1/x.(x+1)=2013/4030
1/2-1/3+1/3-1/4+...+1/x-1/x+1=2013/4030
1/2-1/x+1=2013/4030
1/x+1=1/2015
=> x+1=2015
x=2014
Vậy x=2014
KN
5 tháng 5 2016
Đặt A=Vế trái
Ta có :
\(A \over 2 \)\(= \)\({1\over 6 } +{1\over 12 }+{1\over 20 }+...+{1\over x(x+1)}\)
=\({1\over 2}-{1\over 3}+{1\over 3}-{1\over 4}+{1\over4}-{1\over 5}+...+{1\over x-1}-{1\over x}+{1\over x}-{1\over x+1}\)
=\({1\over2}-{1\over x+1}\)
Từ đó suy ra: \({1\over2}-{1\over x+1}={2013\over4030}\)
=> x=2014
DT
0
NT
2
TM
20 tháng 6 2016
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\) nên x+1=0
=>x=0-1
=>x-1
DL
0
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2015}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{4030}\)
tự làm tiếp nhé mk ăn cơm đã