Ta có : 1+\(\dfrac{1}{x+2}\) = \(\dfrac{12}{8-x^3}\) (đkxđ x\(\ne\pm2\) )
\(\Leftrightarrow\) \(\dfrac{1}{x+2}\) = \(\dfrac{12}{8-x^3}-1\)
\(\Leftrightarrow\)\(\dfrac{1}{x+2}=\dfrac{12-\left(8-x^3\right)}{8-x^3}\)
\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{x^3+4}{8-x^3}\)
\(\Leftrightarrow8-x^3=\left(x+2\right)\left(x^3+4\right)\)
\(\Leftrightarrow8-x^3=x^4+4x+2x^3+8\)
\(\Leftrightarrow-x^3-x^4-4x-2x^3=8-8\)
\(\Leftrightarrow-x^4-3x^3-4x=0\)
\(\Leftrightarrow-x\left(x^3+3x^2+4\right)=0\)
\(\Rightarrow-x=0\)\(\Rightarrow x=0\) (TM x\(\ne\pm2\))
