`x^2(x^2-9)+40=20`
`<=>x^4-9x^2+20=0`
`<=>x^4-4x^2-5x^2+20=0`
`<=>x^2(x^2-4)-5(x^2-4)=0`
`<=>(x^2-5)(x^2-4)=0`
`<=>` \(\left[ \begin{array}{l}x^2-4=0\\x^2-5=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=±2\\x=±\sqrt{5}\end{array} \right.\)
Vậy `S={2;-2;\sqrt{5};-\sqrt{5}}`
`frac{2x}{x-3}=frac{x^2+11x-6}{x^2-9}`
Điều kiện: `x\ne±3`
`<=>frac{2x(x+3)}{(x-3)(x+3)}=frac{x^2+11x-6}{(x-3)(x+3)}`
`=>2x(x+3)=x^2+11x-6`
`<=>2x^2+6x=x^2+11x-6`
`<=>x^2-5x+6=0`
`<=>x^2-3x-2x+6=0`
`<=>x(x-3)-2(x-3)=0`
`<=>(x-3)(x-2)=0`
`<=>` \(\left[ \begin{array}{l}x-3=0\\x-2=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=3(\text{ktmđk})\\x=2(\text{tmđk})\end{array} \right.\)
Vậy `S={2}`
