ĐKXĐ: \(x\ne-2\).
\(PT\Leftrightarrow\dfrac{12}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{x+3}{x+2}\)
\(\Rightarrow12=\left(x+3\right)\left(x^2-2x+4\right)\)
\(\Leftrightarrow x^3+x^2-2x+12=12\)
\(\Leftrightarrow x\left(x^2+x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=1\left(TM\right)\\x=-2\left(l\right)\end{matrix}\right.\).
Vậy x = 0; x = 1.
ĐK: ` x \ne -2`
`12/(8+x^3)=1+1/(x+2)`
`<=>1/((x+2)(x^2-2x+4))` = 1+1/(x+2)`
`<=> 12=x^3 +8+x^2-2x+4`
`<=> x^3+x^2-2x=0`
`<=> x(x^2+x-2)=0`
`<=>` \(\left[{}\begin{matrix}x=0\\x=1\\x=-2\left(L\right)\end{matrix}\right.\)
Vậy `S={0;1}`.
