Câu 1 (2điểm): Giải các phương trình sau. a) $3x-24 = 0$ b) $\frac{x+6}{5}$ - $\frac{x-2}{3}$= 2  c) $\frac{x^2-2}{x^2-4}$ - $\frac{x}{2x-4}$ = $\frac{1}{x+2}$  d) $|1-5x| -16=3$

Đáp án:

$a) x=8\\ b) x=-1\\ c)x=0;x=4\\ d)x =-\dfrac{18}{5};x =4.$

Giải thích các bước giải:

$a) 3x-24=0\\ \Leftrightarrow 3x=24\\ \Leftrightarrow x=8$

Vậy $x=8$

$b) \dfrac{x+6}{5} -\dfrac{x-2}{3} =2\\ \Leftrightarrow 15.\dfrac{x+6}{5} -15.\dfrac{x-2}{3} =15.2\\ \Leftrightarrow 3(x+6) -5(x-2)=30\\ \Leftrightarrow 28-2x=30\\ \Leftrightarrow -2x=2\\ \Leftrightarrow x=-1$

Vậy $x=-1$

$c) \dfrac{x^2-2}{x^2-4} - \dfrac{x}{2x-4} =\dfrac{1}{x+2} \\ \text{ĐKXĐ: } \left\{\begin{array}{l} x^2-4 \ne 0\\2x-4 \ne 0 \\ x+2 \ne 0 \end{array} \right.\Leftrightarrow x \ne \pm 2\\ \dfrac{x^2-2}{x^2-4} - \dfrac{x}{2x-4} =\dfrac{1}{x+2} \\ \Leftrightarrow \dfrac{x^2-2}{(x-2)(x+2)} - \dfrac{x}{2(x-2)} - \dfrac{1}{x+2} =0\\ \Leftrightarrow \dfrac{2(x^2-2)-x(x+2)-2(x-2)}{2(x-2)(x+2)} =0\\ \Leftrightarrow \dfrac{x^2 - 4 x}{2(x-2)(x+2)} =0\\ \Leftrightarrow \dfrac{x(x - 4)}{2(x-2)(x+2)} =0\\ \Rightarrow x(x - 4)=0\\ \Rightarrow x=0;x=4(TM)$

Vậy $x=0;x=4$

$d) |1-5x| -16=3\\ \Leftrightarrow |1-5x| =19\\ \Leftrightarrow \left[\begin{array}{l} 1-5x =19\\ 1-5x =-19\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} -5x =18\\ -5x =-20\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x =-\dfrac{18}{5}\\ x =4\end{array} \right.$

Vậy $x =-\dfrac{18}{5};x =4.$