Решите уравнение [latex] \frac{x}{x+2} + \frac{x+3}{x-2} = \frac{8}{ x^{2} -4} [/latex]

[latex] \frac{x}{x+2} + \frac{x+3}{x-2} = \frac{8}{x^{2}-4} [/latex] ОДЗ: x+2≠0, x-2≠0, x²-4≠0 ⇒  x≠2, x≠-2 [latex] \frac{x(x-2)+(x+3)(x+2)}{(x+2)(x-2)} = \frac{8}{x^{2}-4} \\ \frac{x^{2}-2x+x^{2}+2x+3x+6}{x^{2}-4} = \frac{8}{x^{2}-4} \\ 2x^{2}+3x+6-8=0[/latex] 2x²+3x-2=0 D=9+4*2*2=9+16=25 x1=(-3-5)/4=-8/4=-2 x2=(-3+5)/4=2/4=1/2