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![](https://rs.olm.vn/images/avt/0.png?1311)
Quy đồng vế trái ta có
\(\frac{4026}{x^4+x^2+1}=\frac{2014}{x.\left(x^4+x^2+1\right)}\)
Lại quy đồng 2 vế ta được
\(\frac{4026.x}{x.\left(x^4+x^2+1\right)}=\frac{2014}{x.\left(x^4+x^2+1\right)}\)
Suy ra: 4026.x =2014
<=>\(x=\frac{2014}{4026}\)
rút gọn là xong.OK?
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(x^4+2013x^2+2012x+2013=x^4+2013x^2+2013x+2013-x\)
\(=\left(x^4-x\right)+\left(2013x^2+2013x+2013\right)\)
\(=x\left(x^3-1\right)+2013\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2013\left(x^2+x+1\right)\)
\(=\left(x^2-x+2013\right)\left(x^2+x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(x^4+2013x^2+2012x+2013\)
\(=x^4+2013x^2-x+2013x+2013\)
\(=\left(x^4-x\right)+\left(2013x^2+2013x+2013\right)\)
\(=x\left(x^3-1\right)+2013\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2013\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left\{x\left(x-1\right)+2013\right\}\)
\(=\left(x^2+x+1\right)\left(x^2-x+2013\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)