b)
[tex]3(1-x)+\frac{1}{2}x = \frac{3}{4}(x+6)+5 \ \ /\cdot4\\\\12(1-x)+2x=3(x+6)+20\\\\12-12x+2x=3x+18+20\\\\-10x-3x = 38-12\\\\-13x = 26 \ \ /:(-13)\\\\\boxed{x = -2}[/tex]
c)
[tex]\sqrt{3}x+4 = x-8\\\\\sqrt{3}x-x = -8-4\\\\x(\sqrt{3}-1) = -12 \ \ /:(\sqrt{3}-1)\\\\x = \frac{-12}{\sqrt{3}-1}\cdot\frac{\sqrt{3}+1}{\sqrt{3}+1}\\\\x = \frac{-12(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}}\\\\x = \frac{-12(\sqrt{3}+1)}{3-1}\\\\x = -6(\sqrt{3}+1)\\\\\boxed{x = -6\sqrt{3}-6}[/tex]
d)
[tex]\frac{2x-1}{x+2} = \frac{2x-7}{x-2}\\\\Dziedzina:\\x \neq -2 \ \ i \ \ x\neq 2\\D = R\setminus\{-2, 2\}\\\\(2x-7)(x+2) = (2x-1)(x-2)\\\\2x^{2}+4x-7x-14 = 2x^{2}-4x-x+2\\\\-3x-14=-5x+2\\\\-3x+5x = 2+14\\\\2x = 16 \ \ /:2\\\\\boxed{x = 8}[/tex]
e)
[tex]\frac{(x+3)(x-7)}{2x-14}=0\\\\Dziedzina:\\2x-14\neq 0 \ \ \rightarrow \ \ x \neq 7\\D = R \setminus\{7\}\\\\(x+3)(x-7) = 0\\\\x+3 = 0 \ \vee \ x-7 = 0\\\\x = -3 \ \vee \ x = 7 \ \notin D\\\\\boxed{x = -3}[/tex]
