4/x^2-4+1/(x^2+5x+6)=-5/4

`4/(x^2-4+1)+1/(x^2+5x+6)=-5/4`

ĐKXĐ : $\begin{cases} x^2-4 \ne 0\\x^2+5x+6 \ne 0 \end{cases}$ `<=>` $\begin{cases} x \ne \pm 2\\x \ne -3 \end{cases}$

`<=>` `(4)/((x-2)(x+2))+(1)/((x+2)(x+3))=(-5)/4`

`=>4(x+3)+x-2=(-5)/(4).(x^2+5x+6)(x-2)`

`<=>5x+10=(-5)/4-15/4.x^2+5x+15`

`<=>(-5)/4x^3-15/4x^2+15=0`

`<=>x≈1,61`