(x^2-4)(3x+1/2)=(1/3-1/4-1/12)(2016/2017-2017/2018-2018/2019)

Đáp án+Giải thích các bước giải:

`(x^2 -4)(3x+1/2)=(1/3-1/4-1/12)(2016/2017-2017/2018-2018/2019)`

`<=>(x^2 -4)(3x+1/2)=(4/12-3/12-1/12)(2016/2017-2017/2018-2018/2019)`

`<=>(x^2 -4)(3x+1/2)=0.(2016/2017-2017/2018-2018/2019)`

`<=>(x^2 -4)(3x+1/2)=0`

`<=>` \(\left[ \begin{array}{l}x^2 -4=0\\3x+\dfrac{1}{2}=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=2\\x=-2\\x=\dfrac{-1}{6}\end{array} \right.\) 

Vậy `x\in {2;-1;-1/6}`