Bài 1 (2,0 điểm). Giải các phương trình sau: a. $4x+12$=$0$ b. $(3x+7) - (2x-4)$ = $0$ c.$\frac{x+2}{x.(x-2)}$ -$\frac{1}{x}$ $\frac{4}{x-2}$  d.$|4x|$ = $2x+12$

Đáp án:

$a) x=-3\\ b)x=-\dfrac{7}{3};x=2\\ c)x=1\\ d)x=-2;x=6.$

Giải thích các bước giải:

$a) 4x+12=0\\ \Leftrightarrow  4x=-12\\ \Leftrightarrow x=-3$

Vậy $x=-3$

$b) (3x+7)(2x-4)=0\\ \Leftrightarrow  \left[\begin{array}{l} 3x+7=0\\2x-4=0\end{array} \right.\\ \Leftrightarrow  \left[\begin{array}{l} 3x=-7\\2x=4\end{array} \right.\\ \Leftrightarrow  \left[\begin{array}{l} x=-\dfrac{7}{3}\\x=2\end{array} \right.$

Vậy $x=-\dfrac{7}{3};x=2$

$c) \dfrac{x+2}{x(x-2)}-\dfrac{1}{x}=\dfrac{4}{x-2}\\ \text{ĐK: }x\ne 0; x \ne 2\\ \dfrac{x+2}{x(x-2)}-\dfrac{1}{x}=\dfrac{4}{x-2}\\ \Leftrightarrow \dfrac{x+2}{x(x-2)}-\dfrac{x-2}{x(x-2)}=\dfrac{4x}{x(x-2)}\\ \Leftrightarrow \dfrac{x+2}{x(x-2)}-\dfrac{x-2}{x(x-2)}-\dfrac{4x}{x(x-2)}=0\\ \Leftrightarrow \dfrac{x+2-(x-2)-4x}{x(x-2)}=0\\ \Leftrightarrow \dfrac{4 - 4 x}{x(x-2)}=0\\ \Rightarrow 4-4x=0\\ \Leftrightarrow 4x=4\\ \Leftrightarrow x=1 (TM)$

Vậy $x=1$

$d) |4x|=2x+12\\ \Leftrightarrow \left\{\begin{array}{l} 2x+12 \ge 0\\ \left[\begin{array}{l} 4x=2x+12 \\ -4x=2x+12\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 2x \ge -12\\ \left[\begin{array}{l} 2x=12 \\ -6x=12\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge -6\\ \left[\begin{array}{l} x=6 \\ x=-2\end{array} \right.\end{array} \right.\\ \Leftrightarrow  \left[\begin{array}{l} x=6 \\ x=-2\end{array} \right.$

Vậy $x=-2;x=6.$