Cách giải:
$\dfrac{6}{1-x}=x+4$
$ĐKXĐ:x \neq 1$
$pt \leftrightarrow 6=(x+4)(1-x)$
$\to (x+4)(x-1)=-6$
$\to x^2+3x-4=-6$
$\to x^2+3x+2=0$
$\to x^2+x+2x+2=0$
$\to x(x+1)+2(x+1)=0$
$\to (x+1)(x+2)=0$
$\to \left[ \begin{array}{l}x=-2\\x=-1\end{array} \right.$