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1 tháng 1
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 \(S=\left\{2015\right\}\)
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
\(=\dfrac{x+2014-x}{x\left(x+2014\right)}=\dfrac{2014}{x\left(x+2014\right)}\)
NT
1
21 tháng 12 2018
Đặt biểu thức là A
\(\Rightarrow\)A=\(\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}+\dfrac{\left(x+2\right)-\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}+\dfrac{\left(x+3\right)-\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{\left(x+2014\right)-\left(x+2013\right)}{\left(x+2013\right)\left(x+2014\right)}\)
\(\Leftrightarrow\dfrac{x+1}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}+\dfrac{x+2}{\left(x+1\right)\left(x+2\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{x+2014}{\left(x+2013\right)\left(x+2014\right)}-\dfrac{x+2013}{\left(x+2013\right)\left(x+2014\right)}\)\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}-\dfrac{1}{x+2}-...-\dfrac{1}{x+2013}+\dfrac{1}{x+2013}-\dfrac{1}{x+2014}.\)\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+2014}\)
\(\Leftrightarrow\dfrac{x+2014-x}{x\left(x+2014\right)}\)
\(\dfrac{2014}{x\left(x+2014\right)}\)
16 tháng 12 2019
\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)tương tự những cái kia rồi triệt tiêu còn phân thức đầu vs cuối
25 tháng 1 2023
a: =>\(\left(\dfrac{2x+1}{9}+1\right)+\left(\dfrac{2x+2}{8}+1\right)+...+\left(\dfrac{2x+9}{1}+1\right)=0\)
=>2x+10=0
=>x=-5
b: \(\Leftrightarrow\left(\dfrac{x-1}{2015}-1\right)+\left(\dfrac{x-2}{2014}-1\right)+...+\left(\dfrac{x-2014}{2}-1\right)+\left(x-2016\right)=0\)
=>x-2016=0
=>x=2016
NP
0
21 tháng 7 2018
\(A=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(\Rightarrow A=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2013}-\dfrac{1}{x+2014}\)
\(\Rightarrow A=\dfrac{1}{x}-\dfrac{1}{x+2014}\)
\(\Rightarrow A=\dfrac{2014}{x\left(x+2014\right)}\)
HS
21 tháng 7 2018
\(A=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+....+\dfrac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+...+\dfrac{1}{x+2013}-\dfrac{1}{x+2014}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+2014}-\dfrac{x+2014}{x\left(x+2014\right)}-\dfrac{x}{x\left(x+2014\right)}\)
\(=\dfrac{x+2014-x}{x\left(x+2014\right)}\)
\(=\dfrac{2014}{x\left(x+2014\right)}\)
\(A=\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+.....+\frac{1}{\left(x+2013\right)\left(x+1014\right)}\)
\(\Leftrightarrow A=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+....+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(\Leftrightarrow A=\frac{1}{x}-\frac{1}{x+2014}\)
\(\Leftrightarrow A=\frac{x+2014-x}{x\left(x+2014\right)}=\frac{2014}{x\left(x+2014\right)}\)