Đáp án:
$a) \text{ĐKXĐ: } \left\{\begin{array}{l} x \ne -1 \\ x \ne 3\end{array} \right.\\ b) x=-6$
Giải thích các bước giải:
$a) P=\dfrac{3x^2+3x}{(x+1)(2x-6)}\\ \text{ĐKXĐ: }(x+1)(2x-6) \ne 0\\ \Leftrightarrow \left\{\begin{array}{l} x+1 \ne 0 \\ 2x-6 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ne -1 \\ 2x \ne 6\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ne -1 \\ x \ne 3\end{array} \right.\\ b) P=1\\ \Leftrightarrow \dfrac{3x^2+3x}{(x+1)(2x-6)}=1\\ \Leftrightarrow \dfrac{3x(x+1)}{(x+1)(2x-6)}=1\\ \Leftrightarrow \dfrac{3x}{2x-6}=1\\ \Leftrightarrow 3x=2x-6\\ \Leftrightarrow x=-6$
Vậy với $x=-6$ thì $P=1.$
