Решите неравенствоа)[latex] \frac{ x^{2}-6x+4 }{x-1} больше 0[/latex]б)[latex] \frac{3- x^{2} }{3 x^{2}-4x-1} больше 0[/latex]

[latex] \frac{x^{2}-6x+4}{x-1}>0, \\ x^2-6x+4=0, D_1=5, \\ x_1=3-\sqrt{5}, x_2=3+\sqrt{5}, \\ x^2-6x+4=(x+\sqrt{5})(x-\sqrt{5}); \\ x-1=0, x=1; \\ \frac{(x+\sqrt{5})(x-\sqrt{5})}{x-1}>0, \\ (x+\sqrt{5})(x-1)(x-\sqrt{5})>0, \\ x\in(-\sqrt{5};1)\cup(\sqrt{5};+\infty).[/latex] [latex] \frac{3- x^{2}}{3 x^{2}-4x-1}>0, \\ 3-x^2=0, \\ x^2=3, \\ x_1=-\sqrt{3}, x_2=\sqrt{3}, \\ 3-x^2=-(x+\sqrt{3})(x-\sqrt{3}), \\ 3x^2-4x-1=0, \\ D_1=7, \\ x_1=\frac{2-\sqrt{7}}{3} , x_2=\frac{2+\sqrt{7}}{3}, \\ 3x^2-4x-1=3(x-\frac{2-\sqrt{7}}{3})(x-\frac{2+\sqrt{7}}{3}); \\ \frac{-(x+\sqrt{3})(x-\sqrt{3})}{3(x-\frac{2-\sqrt{7}}{3})(x-\frac{2+\sqrt{7}}{3})}>0, \\ (x+\sqrt{3})(x-\frac{2-\sqrt{7}}{3})(x-\frac{2+\sqrt{7}}{3})(x-\sqrt{3})<0, \\ [/latex] [latex]x\in(-\sqrt{3},\frac{2-\sqrt{7}}{3})\cup(\frac{2+\sqrt{7}}{3};\sqrt{3})[/latex]