ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2014};-\dfrac{2}{2015};-\dfrac{3}{2016};-\dfrac{4}{2017}\right\}\)

Ta có: \(\dfrac{1}{2014x+1}-\dfrac{1}{2015x+2}=\dfrac{1}{2016x+3}-\dfrac{1}{2017x+4}\)

\(\Leftrightarrow\dfrac{2015x+2-2014x-1}{\left(2014x+1\right)\left(2015x+2\right)}=\dfrac{2017x+4-2016x-3}{\left(2016x+3\right)\left(2017x+4\right)}\)

\(\Leftrightarrow\dfrac{x+1}{\left(2014x+1\right)\left(2015x+2\right)}-\dfrac{x+1}{\left(2016x+3\right)\left(2017x+4\right)}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{\left(2014x+1\right)\left(2015x+2\right)}-\dfrac{1}{\left(2016x+3\right)\left(2017x+4\right)}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\\dfrac{1}{\left(2014x+1\right)\left(2015x+2\right)}=\dfrac{1}{\left(2016x+3\right)\left(2017x+4\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\4058210x^2+6043x+2=4066272x^2+14115x+12\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\8062x^2+8072x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\8062x^2+8062x+10x+10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\8062x\left(x+1\right)+10\left(x+1\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\\left(x+1\right)\left(8062x+10\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x+1=0\\8062x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-1\\8062x=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(nhận\right)\\x=\dfrac{-5}{4031}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{-1;\dfrac{-5}{4031}\right\}\)

thanks

       \(x^4+2015x^2+2014x+2015=0\)

\(\Leftrightarrow\)\(\left(x^4+x^2+1\right)+\left(2014x^2+2014x+2014\right)=0\)

\(\Leftrightarrow\)\(\left(x^2+x+1\right)\left(x^2-x+1\right)+2014\left(x^2+x+1\right)=0\)

\(\Leftrightarrow\)\(\left(x^2+x+1\right)\left(x^2-x+2015\right)=0\)

Ta có:   \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

           \(\left(x-\frac{1}{2}\right)^2+2014\frac{3}{4}>0\)

Vậy  pt  vô nghiệm

ai làm hộ mk với 

tks nhiều

\(\frac{2}{x^2-2015x+2014}=\frac{1}{x^2-2016x+2015}\)

\(\Leftrightarrow\frac{2}{\left(x-1\right)\left(x-2014\right)}=\frac{1}{\left(x-1\right)\left(x-2015\right)}\)

\(\Leftrightarrow\frac{2}{x-2014}=\frac{1}{x-2015}\)

áp dụng tính chất tỉ lệ thức ta có:

\(\frac{2}{x-2014-2}=\frac{1}{x-2015-1}\)

\(\Leftrightarrow\frac{2}{x-2016}-\frac{1}{x-2016}=0\)

\(\Leftrightarrow\left(x-2016\right)\left(2-1\right)=0\)

\(\Leftrightarrow x-2016=0\)

\(\Leftrightarrow x=2016\)

mk nhầm đề sửa lại là phân tích đa thức thành nhân tử

x⁴+2014x²-2014x-x+2014

=x(x³-1)+2014(x²-x-1)

=x(x-1)(x²-x-1)+2014(x²-x-1)

=(x²-x-1)(x²-x+2014)

em gửi bài qua fb thầy chữa cho, tìm fb của thầy bằng sđt nhé: 0975705122

(2015x - 2014)³ = 8(x - 1)³ + (2013x - 2012)³

<=> 6(x - 1)(2013x - 2012)(2015x - 2014) = 0

Tới đây thì xong rồi

em gửi bài qua fb thầy chữa cho nhé, tìm fb của thầy bằng sđt: 0975705122 nhé.