\(\dfrac{1}{x+2}+\dfrac{6x+12}{x^3+8}-\dfrac{7}{x^2-2x+4}=0\) \(\left(đk:x\ne-2\right)\)
\(\Leftrightarrow\dfrac{x^2-2x+4+6x+12-7\left(x+2\right)}{x^3+8}=0\)
\(\Leftrightarrow\dfrac{x^2-3x+2}{x^3+8}=0\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)(TM)
Vậy ...
dk : x khac -2
\(\Rightarrow x^2-2x+4+6x+12-7\left(x+2\right)=0\)
\(\Leftrightarrow x^2+4x+16-7x-14=0\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow x^2-2x-x+2=0\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\Leftrightarrow x=1;x=2\)