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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: \(S=\left\{-1;\dfrac{-5}{4031}\right\}\)
\(x^4+2015x^2+2014x+2015=0\)
\(\Leftrightarrow\)\(\left(x^4+x^2+1\right)+\left(2014x^2+2014x+2014\right)=0\)
\(\Leftrightarrow\)\(\left(x^2+x+1\right)\left(x^2-x+1\right)+2014\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\)\(\left(x^2+x+1\right)\left(x^2-x+2015\right)=0\)
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
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)
\(\frac{2}{x^2-2015x+2014}=\frac{1}{x^2-2016x+2015}\)
\(\Leftrightarrow\frac{2}{\left(x-1\right)\left(x-2014\right)}=\frac{1}{\left(x-1\right)\left(x-2015\right)}\)
\(\Leftrightarrow\frac{2}{x-2014}=\frac{1}{x-2015}\)
áp dụng tính chất tỉ lệ thức ta có:
\(\frac{2}{x-2014-2}=\frac{1}{x-2015-1}\)
\(\Leftrightarrow\frac{2}{x-2016}-\frac{1}{x-2016}=0\)
\(\Leftrightarrow\left(x-2016\right)\left(2-1\right)=0\)
\(\Leftrightarrow x-2016=0\)
\(\Leftrightarrow x=2016\)
b: \(x^4+x^2+1\)
\(=x^4+2x^2+1-x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
c: \(x^7+x^5+1\)
\(=x^7+x^6+x^5-x^6-x^5-x^4+x^5+x^4+x^3-x^3+1\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)
= 1/2011x-1/2012x+1+1/2014x+1=1/2013x+1
đặt 2011x+1=a; 2012x+1=b; 2014x+1=c Ta có
1/a+1/b+1/c=1/a+b+c *Tự cm nhé!*
= a=-b hoặc b=-c hoặc c=-a
* Nếu a=-b =>2011x+1=-2012x-1=>x=..... tính ra
*Nếu b=-c => 2012x+1=-2014x-1=> x=....
*Nếu c=-a => 2014x+1=-2011x-1=> x=...
Vậy.....
ĐKXĐ : \(2014x+1\ne0;2015x+2\ne0;2016x+3\ne0;2017x+4\ne0\)
=> \(x\ne-\frac{1}{2014};x\ne-\frac{2}{2015};x\ne-\frac{3}{2016};x\ne-\frac{4}{2017}\)
Ta có \(\frac{1}{2014x+1}-\frac{1}{2015x+2}=\frac{1}{2016x+3}-\frac{1}{2017x+4}\)
<=> \(\frac{x+1}{\left(2014x+1\right)\left(2015x+2\right)}=\frac{x+1}{\left(2016x+3\right)\left(2017x+4\right)}\)
<=> \(\left(x+1\right)\left(\frac{1}{\left(2014x+1\right)\left(2015x+2\right)}-\frac{1}{\left(2016x+3\right)\left(2017x+4\right)}\right)=0\)
<=> \(\left(x+1\right)\left(\frac{8062x^2+8072x+10}{\left(2014x+1\right)\left(2015x+2\right)\left(2016x+3\right)\left(2017x+4\right)}\right)=0\)
<=> \(\left(x+1\right)\frac{\left(x+1\right)\left(8062x+10\right)}{\left(2014x+1\right)\left(2015x+2\right)\left(2016x+3\right)\left(2017x+4\right)}=0\)
<=> (x + 1)2(8062x + 10) = 0
<=> \(\orbr{\begin{cases}x+1=0\\8062x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\left(tm\right)\\x=-\frac{5}{4031}\left(tm\right)\end{cases}}\)
Vậy \(x\in\left\{-1;-\frac{5}{4031}\right\}\)là nghiệm phương trình