`x/{x+1}+{2x}/{(x+1)(2x+1)}+{3x}/{(x+1)(2x+1)(3x+1)}>1` $(1)$

$ĐK: \begin{cases}x+1\ne 0\\2x+1\ne 0\\3x+1\ne 0\end{cases}$ $⇔\begin{cases}x\ne -1\\x\ne \dfrac{-1}{2}\\x\ne \dfrac{-1}{3}\end{cases}$

`(1)<=>{x(2x+1)(3x+1)+2x(3x+1)+3x-1(x+1)(2x+1)(3x+1)}/{(x+1)(2x+1)(3x+1)}>0`

`<=>{(2x+1)(3x+1)(x-x-1)+2x(3x+1)+3x}/{(x+1)(2x+1)(3x+1)}>0`

`<=>{-(2x+1)(3x+1)+2x(3x+1)+3x}/{(x+1)(2x+1)(3x+1)}>0`

`<=>{(3x+1)(2x-2x-1)+3x}/{(x+1)(2x+1)(3x+1)}>0`

`<=>{-1}/{(x+1)(2x+1)(3x+1)}>0`

`<=>(x+1)(2x+1)(3x+1)<0`

$⇔\left[\begin{array}{l}\begin{cases}(x+1)(2x+1)>0\\3x+1<0\end{cases}\\\begin{cases}(x+1)(2x+1)<0\\3x+1>0\end{cases}\end{array}\right.$

$⇔\left[\begin{array}{l}\begin{cases}x>\dfrac{-1}{2}\ hoặc \ x< -1\\x<\dfrac{-1}{3}\end{cases}\\\begin{cases}-1<x<\dfrac{-1}{2}\\x>\dfrac{-1}{3}\end{cases} (loại) \end{array}\right.$

$⇒\left[\begin{array}{l}\dfrac{-1}{2}<x<\dfrac{-1}{3}\\x< -1\end{array}\right.$

Vậy bất phương trình có nghiệm $x$ thỏa:

`{-1}/2<x<{-1}/3` hoặc `x< -1`