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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
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
LT
3
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
7 tháng 5 2018
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2015}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)
\(1-\frac{1}{x+1}=1-\frac{2015}{2016}\)
\(\frac{1}{x+1}=\frac{1}{2016}\)
\(x=2016-1\)
\(\Rightarrow x=2015\)
27 tháng 4 2018
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
27 tháng 4 2018
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}.\)
<=>\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)
<=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)
<=> \(-\frac{1}{x+1}=\frac{-1}{2016}\) <=> x+1 = 2016 <=> x = 2015
4 tháng 2 2021
a) Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2014}\right)\left(1-\dfrac{1}{2015}\right)\left(1-\dfrac{1}{2016}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2013}{2014}\cdot\dfrac{2014}{2015}\cdot\dfrac{2015}{2016}\)
\(=\dfrac{1}{2016}\)
b) Ta có: \(\dfrac{x-2}{12}+\dfrac{x-2}{20}+\dfrac{x-2}{30}+\dfrac{x-2}{42}+\dfrac{x-2}{56}+\dfrac{x-2}{72}=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\cdot\dfrac{2}{9}=\dfrac{16}{9}\)
\(\Leftrightarrow x-2=\dfrac{16}{9}:\dfrac{2}{9}=\dfrac{16}{9}\cdot\dfrac{9}{2}=8\)
hay x=10
Vậy: x=10
NT
1
20 tháng 4 2016
\(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)
\(1-\frac{1}{x+1}=\frac{2015}{2016}\)
\(\frac{1}{x+1}=1-\frac{2015}{2016}=\frac{1}{2016}\)
\(x+1=2016=>x=2015\)
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
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