a. đkxđ: x khác -2
\(1+\dfrac{1}{x+2}=\dfrac{12}{8+x^3}\)
\(\Leftrightarrow\dfrac{x^3+2^3}{\left(x+2\right)\left(x^2-2x+4\right)}+\dfrac{x^2-2x+4}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{12}{\left(x+2\right)\left(x^2-2x+4\right)}\)
\(\Leftrightarrow x^3+8+x^2-2x+4=12\)
\(\Leftrightarrow x^3+x^2-2x=0\)
\(\Leftrightarrow x^3-x^2+2x^2-2x=0\)
\(\Leftrightarrow\left(x^3-x^2\right)+\left(2x^2-2x\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)+2x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2+2x\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\left(loai\right)\\x=1\end{matrix}\right.\)
Vậy...........
b.
\(\dfrac{4x-8}{2x^2+1}=0\)
\(\Leftrightarrow4x-8=0\)
\(\Leftrightarrow x=2\)
Vậy..........
