1/2+1/6+1/12....+1/x(x+1)=2015/2016
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27 tháng 4 2018
\(\frac{1}{2}\)+\(\frac{1}{6}\)+\(\frac{1}{12}\)+....+\(\frac{1}{x\left(x+1\right)}\)=\(\frac{2015}{2016}\)
\(\frac{1}{1x2}\)+\(\frac{1}{2x3}\)+\(\frac{1}{3x4}\)+.....+\(\frac{1}{x\left(x+1\right)}\)=\(\frac{2015}{2016}\)
\(1\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-.....+\(\frac{1}{x}\)-\(\frac{1}{x+1}\)=\(\frac{2015}{2016}\)
1-\(\frac{1}{x+1}\) = \(\frac{2015}{2016}\)
\(\frac{1}{x+1}\) =1- \(\frac{2015}{2016}\)
\(\frac{1}{x+1}\) = \(\frac{1}{2016}\)
\(\Rightarrow\)x + 1= 2016
\(\Rightarrow\)x = 2015
5 tháng 5 2016
\(\frac{1}{2}+\frac{1}{6}+.....+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)
\(=1-\frac{1}{x+1}=\frac{2015}{2016}\)
\(=\frac{1}{x+1}=\frac{1}{2016}\)
=> x + 1 = 2016
=> x =2015
LD
7 tháng 5 2016
VẾ TRÁI LÀ:
A=1/2.3+1/3.2+1/4.5+...+1/x[x+1]
A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/n-1/n+1
A=1-1/n+1
1-1/n+1=2015/2016
1/n+1=1-2015/2016
1/n+1=1/2016
n=2016-1
n=2015
29 tháng 5 2016
Vế trái là
S=1/2+1/6+1/12+1/20+..+1/x(x+1)
S=1/1.2+1/2.3+1/3.4+1/4.5+...+1/x.(x+1)
S=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/x.(x+1)
S=1-1/(x+1)=Vế phải=2015/2016
<=>1-1/(x+1)=2015/2016
1/(x+1)=1/2016
=>x+1=2016
x=2015
Ủng hộ mk mha Chí Tiến
NT
1
20 tháng 4 2016
\(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)
\(1-\frac{1}{x+1}=\frac{2015}{2016}\)
\(\frac{1}{x+1}=1-\frac{2015}{2016}=\frac{1}{2016}\)
\(x+1=2016=>x=2015\)
DA
2
24 tháng 4 2016
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x.(x+1) = 2015 /2016
1-1/2 + 1/2 - 1/3 + ... + 1/x - 1/x+1=2015/2016
1-1/x+1 =2015/2016
1/x+1 = 1 - 2015/2016
1/x+1 = 1/2016
=> x+1 = 2016
=> x=2016 - 1 = 2015
vậy x = 2015
LT
4
13 tháng 5 2016
1/2+1/6+1/12+1/20+...+1/x(x+1)=2015/2016
1/1.2+1/2.3+1/3.4+.....+1/x.(x+1)=2015/2016
1-1/2+1/2-1/3+1/3-1/4+......+1/x-1/x+1=2015/2016
1-1/x-1=2015/2016
1/x+1=1-2015/2016
1/x+1=1/2016
=> x+1=2016
x=2016-1
x=2015
vậy x =2015
tích mình nha
HP
13 tháng 5 2016
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.......+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)
=>\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)
=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)
=>\(1-\frac{1}{x+1}=\frac{2015}{2016}\)
=>\(\frac{1}{x+1}=1-\frac{2015}{2016}=\frac{1}{2016}\)
=>x+1=2016
=>x=2015
Vậy x=2015
ND
2
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
1/1x2+1/2x3+...+1/x(x+1)=2015/2016
1/1-1/2+1/2-1/3+...+1/x-1/x+1=2015/2016
2/1-1/x+1=2015/2016
2016/2016-1/x+1=2015/2016
1/x+1=2016/2016-2015/2016
1/x+1=1/2016
x+1=2016
x=2016-1
x=2015
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}.\)
<=>\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)
<=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)
<=> \(-\frac{1}{x+1}=\frac{-1}{2016}\) <=> x+1 = 2016 <=> x = 2015