[tex]\dfrac{6}{x}\leq4\qquad(x\not=0)\\\\6x\leq4x^2\\4x^2-6x\geq0\\2x(2x-3)\geq0\\x\in(-\infty,0\rangle\cup\left\langle\dfrac{3}{2},\infty\right)\\\\\\x\in(-\infty,0\rangle\cup\left\langle\dfrac{3}{2},\infty\right) \wedge x\not=0\\\boxed{x\in(-\infty,0)\cup\left\langle\dfrac{3}{2},\infty\right)}[/tex]
