KT
1
K
Khách
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NT
1
21 tháng 12 2018
Đặt biểu thức là A
\(\Rightarrow\)A=\(\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}+\dfrac{\left(x+2\right)-\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}+\dfrac{\left(x+3\right)-\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{\left(x+2014\right)-\left(x+2013\right)}{\left(x+2013\right)\left(x+2014\right)}\)
\(\Leftrightarrow\dfrac{x+1}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}+\dfrac{x+2}{\left(x+1\right)\left(x+2\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{x+2014}{\left(x+2013\right)\left(x+2014\right)}-\dfrac{x+2013}{\left(x+2013\right)\left(x+2014\right)}\)\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}-\dfrac{1}{x+2}-...-\dfrac{1}{x+2013}+\dfrac{1}{x+2013}-\dfrac{1}{x+2014}.\)\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+2014}\)
\(\Leftrightarrow\dfrac{x+2014-x}{x\left(x+2014\right)}\)
\(\dfrac{2014}{x\left(x+2014\right)}\)
TM
1
19 tháng 3 2020
\(A=\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+.....+\frac{1}{\left(x+2013\right)\left(x+1014\right)}\)
\(\Leftrightarrow A=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+....+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(\Leftrightarrow A=\frac{1}{x}-\frac{1}{x+2014}\)
\(\Leftrightarrow A=\frac{x+2014-x}{x\left(x+2014\right)}=\frac{2014}{x\left(x+2014\right)}\)
1 tháng 1 2024
1)
\(\dfrac{x-1}{2014}+\dfrac{x-2}{2013}+\dfrac{x-3}{2012}+...+\dfrac{x-2014}{1}=2014\)
\(\Leftrightarrow\left(\dfrac{x-1}{2014}-1\right)+\left(\dfrac{x-2}{2013}-1\right)+...+\left(\dfrac{x-2014}{1}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2015}{2014}+\dfrac{x-2015}{2013}+...+\dfrac{x-2015}{1}=0\)
\(\Leftrightarrow\left(x-2025\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1}\right)=0\)
\(\Leftrightarrow x=2015\)
Vậy \(S=\left\{2015\right\}\)
NP
0
27 tháng 1 2023
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2013}-\dfrac{1}{x+2014}\)
=1/x-1/x+2014
\(=\dfrac{x+2014-x}{x\left(x+2014\right)}=\dfrac{2014}{x\left(x+2014\right)}\)
29 tháng 12 2015
2) xét tử ta có
2014+2013/2+2012/3+...+2/2013+1/2014
=(1+2013/2)+(1+2012/3)+...+(1+2/2013)+(1+1/2014)+1
=2015/2+2015/3+...+2015/2013+2015/2014+2015/2015
=2015(1/2+1/3+...+1/2013+1/2014+1/2015) (1)
mà mẫu bằng 1/2+1/3+1/4+...+1/2014+1/2015 (2)
từ (1),(2)=> phân thức trên =2015
LT
0
24 tháng 4 2018
\(\frac{1-x}{2013}=1+\frac{2-x}{2012}-\frac{x}{2014}\)
\(\Leftrightarrow\)\(\frac{1-x}{2013}+1=\frac{2-x}{2012}+1-\left(\frac{x}{2014}-1\right)\)
\(\Leftrightarrow\)\(\frac{2014-x}{2013}=\frac{2014-x}{2012}-\frac{x-2014}{2014}\)
\(\Leftrightarrow\)\(\frac{2014-x}{2013}-\frac{2014-x}{2012}+\frac{2014-x}{2014}\)=0
\(\Leftrightarrow\)(2014-x)(\(\frac{1}{2013}-\frac{1}{2012}+\frac{1}{2014}\))=0
\(\Leftrightarrow\)2014-x=0(do 1/2013 -1/2012 -1/2014)
\(\Leftrightarrow\)x=2014
Answer:
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+2013+2014\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(=\frac{1}{x}-\frac{1}{x+2014}\)
\(=\frac{x+2014-x}{x\left(x+2014\right)}\)
\(=\frac{2014}{x\left(x+2014\right)}\)