`x^2+[-18]/[x^2+x]=3-x` `ĐK: x \ne -1,x \ne 0`
`<=>[x^2(x^2+x)-18]/[x^2+x]=[(3-x)(x^2+x)]/[x^2+x]`
`=>x^4+x^3-18=3x^2+3x-x^3-x^2`
`<=>x^4+2x^3-2x^2-3x-18=0`
`<=>x^4-2x^3+4x^3-8x^2+6x^2-12x+9x-18=0`
`<=>x^3(x-2)+4x^2(x-2)+6x(x-2)+9(x-2)=0`
`<=>(x-2)(x^3+4x^2+6x+9)=0`
`<=>(x-2)(x^3+3x^2+x^2+3x+3x+9)=0`
`<=>(x-2)[x^2(x+3)+x(x+3)+3(x+3)]=0`
`<=>(x-2)(x+3)(x^2+x+3)=0`
`<=>` $\left[\begin{matrix} x=2 (t/m)\\ x=-3 (t/m)\\x^2+x+3=0\text{ (Vô nghiệm)}\end{matrix}\right.$
Vậy `S={-3;2}`
\(x^2+\dfrac{-18}{x^2+x}=3-x\)
\(\Leftrightarrow x^2-\dfrac{18}{x\left(x+1\right)}=3-x\);\(ĐK:x\ne0;-1\)
\(\Leftrightarrow-\dfrac{18}{x\left(x+1\right)}=3-x-x^2\)
\(\Leftrightarrow\dfrac{18}{x\left(x+1\right)}=x^2+x-3\)
\(\Leftrightarrow\dfrac{18}{x\left(x+1\right)}=x\left(x+1\right)-3\)
Đặt \(x\left(x+1\right)=a\)
\(\Leftrightarrow\dfrac{18}{a}=a-3\)
\(\Leftrightarrow a^2-3a-18=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-3\end{matrix}\right.\)
Với `x=6`
`=>`\(x^2+x=6\)
`<=>x^2+x-6=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\) \((tm)\)
Với `x=-3`
`=>`\(x^2+x=-3\)
`<=>x^2+x+3=0` ( vô lý )
Vậy \(S=\left\{2;-3\right\}\)
