`a) 1/(x+1) -4/(x^2-x+1)=(2x^2+1)/(x^3+1) `
`<=> 1/(x+1) -4/(x^2-x+1)-(2x^2+1)/(x^3+1)=0 `
`<=> (x^2-x+1)/((x+1)(x^2-x+1)) -(4(x+1))/((x^2-x+1)(x+1))-(2x^2+1)/((x+1)(x^2-x+1))=0 `
`<=> ((x^2-x+1)-(4x+4)-(2x^2+1))/((x+1)(x^2-x+1))=0 `
`<=> (x^2-x+1-4x-4-2x^2-1)/((x+1)(x^2-x+1))=0 `
`<=> (-x^2-5x-4)/((x+1)(x^2-x+1))=0 `
$⇔\begin{cases}-x^2-5x-4=0\\(x+1)(x^2-x+1)\ne0\end{cases}$
$⇔\begin{cases}-(x^2+5x+4)=0\\x\ne-1\end{cases}$
$⇔\begin{cases}x^2+x+4x+4=0\\x\ne-1\end{cases}$
$⇔\begin{cases}x(x+1)+4(x+1)=0\\x\ne-1\end{cases}$
$⇔\begin{cases}(x+1)(x+4)=0\\x\ne-1\end{cases}$
$⇔\begin{cases}\left[ \begin{array}{l}x=-1(KTM)\\x=-4\end{array} \right.\\x\ne-1\end{cases}$\(\)
`to x=-4`
`b)( 3x+1)/(x+1) -(2x-5)/(x-3)=1 -4/((x+1)(x-3))`
`<=>( 3x+1)/(x+1) -(2x-5)/(x-3)=((x+1)(x-3) -4)/((x+1)(x-3))`
`<=>( 3x+1)/(x+1) -(2x-5)/(x-3)=(x^2+2x-3 -4)/((x+1)(x-3))`
`<=>( 3x+1)/(x+1) -(2x-5)/(x-3)=(x^2+2x-7)/((x+1)(x-3))`
`<=>( 3x+1)/(x+1) -(2x-5)/(x-3)-(x^2+2x-7)/((x+1)(x-3))=0`
`<=>( (3x+1)(x-3))/((x+1)(x-3)) -((2x-5)(x+1))/((x-3)(x+1))-(x^2+2x-7)/((x+1)(x-3))=0`
`<=>( 3x^2-8x-3-(2x^2-3x-5)-(x^2-2x+7))/((x+1)(x-3))=0`
`<=>( 3x^2-8x-3-2x^2+3x+5-x^2+2x-7)/((x+1)(x-3))=0`
`<=>(-3x+5)/((x+1)(x-3))=0`
`<=>`$\begin{cases}-3x+5=0\\(x+1)(x-3)\ne0\end{cases}$
`<=>`$\begin{cases}x=\cfrac{5}{3}\\\left[ \begin{array}{l}x\ne-1\\x\ne3\end{array} \right.\end{cases}$
