Cho A = {x thuộc R |trị tuyệt đối (mx−3) = mx−3}, B = {x thuộc R |x^2−4 = 0

* Hướng dẫn giải

Đáp án C

Ta có: $x\in A\iff mx-3\ge 0$

+ Nếu m = 0 thì $-3\ge 0$ (vô lí) nên $A=\varnothing$

+ Nếu m>0 thì $x\ge \frac{3}{m}$ hay $A=\left[\frac{3}{m};+\infty \right)$

+ Nếu m<0 thì $x\le \frac{3}{m}$ hay $A=\left(-\infty ;\frac{3}{m}\right]$

$x\in B\iff \left[\begin{array}{c}x=2\\ x=-2\end{array}\right.$ hay $B=\left\{-2;2\right\}$

Ta có:

$$\begin{gathered} B\backslash A=B\iff B\cap A=\varnothing \\ \iff \left[\begin{array}{c}m=0\\ \left\{\begin{array}{c}m>0\\ \frac{3}{m}>2\end{array}\right.\\ \left\{\begin{array}{c}m<0\\ \frac{3}{m}<-2\end{array}\right.\end{array}\right.\iff \left[\begin{array}{c}m=0\\ 0<m<\frac{3}{2}\\ -\frac{3}{2}<m<0\end{array}\right. \\ \iff -\frac{3}{2}<m<\frac{3}{2} \end{gathered}$$