`(x^2+x-5)/x+(3x)/(x^2+x-5)+4=0` (ĐKXD: x $\neq$ 0;$\frac{21}{4}$ )
⇒`(x^2+x-5)^2+3x^2+4x(x^2+x-5)=0`
⇔`(x^2)^2+x^2+5^2+2x^3-10x-10x^2+3x^2+4x^3+4x^2-20x=0`
⇔`x^4+6x^3-2x^2-30x+25=0`
⇔`x^4-x^3+7x^3-7x^2+5x^2-5x-25x+25=0`
⇔`x^3(x-1)+7x^2(x-1)+5x(x-1)-25(x-1)=0`
⇔`(x^3+7x^2+5x-25)(x-1)=0`
⇔`(x^3+5x^2+2x^2+10x-5x-25)(x-1)=0`
⇔`[x^2(x+5)+2x(x+5)-5(x+5)](x-1)=0`
⇔`(x^2+2x-5)(x+5)(x-1)=0`
`TH1: x^2+2x-5=0`
⇔`(x^2+2x-1)-4=0`
⇔`(x-1)^2=4`
⇔\(\left[ \begin{array}{l}x-1=2\\x-1=-2\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.\)
`TH2:x+5=0`
⇔`x=-5`
`TH3:x-1=0`
`x=1`
Vậy `S∈{-5;±-1;3}`
