Ответ:

ОДЗ

х-3 ≠0

х≠3

х+4≠0

х≠-4

[tex]\frac{x+4}{x-3} - \frac{x-3}{x+4} = \frac{3}{2} \\\frac{x+4}{x-3} = t \\t-\frac{1}{t} = \frac{3}{2} \\\frac{t^2-1}{t} = \frac{3}{2} \\t^2-1= 1.5t\\t^2-1.5t-1 = 0\\D = 2.25+4 = 6.25 = 2.5^2\\t1 = (1.5-2.5):2 = -1:2 = -0.5\\t2= (1.5+2.5):2 = 4:2 = 2\\\\\frac{x+4}{x-3} = 2\\x+4 = 2x-6\\x = 10\\\\\frac{x+4}{x-3} = -0.5 \\x+4 = -0.5x+1.5\\1.5x = -2.5\\x = -\frac{5}{3}[/tex]