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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
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
\(\Leftrightarrow x+2016=0\).Do \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\)
\(\Leftrightarrow x=-2016\)
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
\(\Leftrightarrow x+2016=0\).Do \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\)
\(\Leftrightarrow x=-2016\)
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\)\(\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(\Leftrightarrow\)\(\frac{x+1}{2015}+\frac{2015}{2015}+\frac{x+20}{2014}+\frac{2014}{2014}=\frac{x+3}{2013}+\frac{2013}{2013}+\frac{x+4}{2012}+\frac{2012}{2012}\)
\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\Leftrightarrow\)\(\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
Suy ra \(x+2016=0\) \(\Leftrightarrow x=-2016\)
Vậy \(x\in\left\{-2016\right\}\)
bạn cộng thêm 1 vào mỗi phân thức đó
sau đó sẽ có phân tử chung là x+2016
kết quả là x=-2016
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\left(\frac{x+1}{2015}+1\right)+\left(\frac{x+2}{2014}+1\right)=\left(\frac{x+3}{2013}+1\right)+\left(\frac{x+4}{2012}+1\right)\)
\(\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\left(x+2016\right).\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
\(x+2016=0\)
\(x=-2016\)
Đặt \(\frac{x}{2013}=\frac{y}{2014}=\frac{z}{2015}=k\)
\(\Rightarrow\hept{\begin{cases}x=2013k\\y=2014k\\z=2015k\end{cases}}\)
Ta có :
4(x - y)(y - z) = 4(2013k - 2014k)(2014k - 2015k)
=4.(-k).(-k) = 4k2 (1)
(z - x)2 = (2015k - 2013k)2 = (2k)2 = 4k2 (2)
Từ 1 và 2
=> 4(x - y)(y - z) = (z - x)2
Theo bài ra , ta có :
\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\left(\frac{x+2}{2014}+1\right)+\left(\frac{x+1}{2015}+1\right)=\left(\frac{x+3}{2013}+1\right)+\left(\frac{x+4}{2012}+1\right)\)
\(\Leftrightarrow\left(\frac{x+2+2014}{2014}\right)+\left(\frac{x+1+2015}{2015}\right)=\left(\frac{x+3+2013}{2013}\right)+\left(\frac{x+4+2012}{2012}\right)\)
\(\Leftrightarrow\frac{x+2016}{2014}+\frac{x+2016}{2015}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\Leftrightarrow\frac{x+2016}{2014}+\frac{x+2016}{2015}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
Vì \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)>0\)
\(\Leftrightarrow x+2016=0\)
\(\Leftrightarrow x=-2016\)
Vậy \(x=-2016\)
Tập nghiệm của phương trình là \(S=\left\{-2016\right\}\)
Chúc bạn học tốt =))
\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\frac{x+2}{2014}+1+\frac{x+1}{2015}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(\frac{x+2+2014}{2014}+\frac{x+1+2015}{2015}=\frac{x+3+2013}{2013}+\frac{x+4+2012}{2012}\)
\(\frac{x+2016}{2014}+\frac{x+2016}{2015}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\frac{x+2016}{2014}+\frac{x+2016}{2015}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\left(x+2016\right).\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
MÀ \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
\(A=B\)
\(\Leftrightarrow\)\(\frac{x+1}{2015}+\frac{x+2}{2014}+\frac{x+3}{2013}+\frac{x+4}{2012}=-2^2\)
\(\Leftrightarrow\)\(\frac{x+1}{2015}+1+\frac{x+2}{2014}+1+\frac{x+3}{2013}+1+\frac{x+4}{2012}+1=0\)
\(\Leftrightarrow\)\(\frac{x+2016}{2015}+\frac{x+2016}{2014}+\frac{x+2016}{2013}+\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\)\(\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}\right)=0\)
\(\Leftrightarrow\)\(x+2016=0\) (do 1/2015 + 1/2014 + 1/2013 + 1/2012 # 0)
\(\Leftrightarrow\)\(x=-2016\)
Vậy...
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
Có: \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)