[tex]a) \frac{3}{x^{2} -4} D=x^{2} -4\neq 0, x\neq 2, x\neq -2\\D=R/[-2,2]\\b) \frac{3x-2}{(2x-4)(x+4)(x-5)} = D=2x-4\neq 0, x\neq 2, x-5\neq 0, x\neq 5\\D=R/[-5,2,5][/tex]
[tex]a) \frac{4x}{x+7} =\frac{5}{x-2} \\D=R/[-7,2]\\4x(x-2)=5(x+7)\\4x(x-2)-5(x+7)=0\\4x^{2} -8x-5x-35=0\\4x^{2} -13x-35=0\\delta= (-13)^{2} -4*4*(-35)=729\\\sqrt{delta} = 27\\x_{1} = \frac{13-27}{8} = -\frac{7}{4} \\x_{2} = \frac{13+27}{8} = 5[/tex][tex]b) \frac{5x-8}{x^{2} -25} =0\\D=R/[-5,5]\\5x-8=0\\5x=8\\x=\frac{8}{5}[/tex]
