Question

помогите решить,даю 100 баллов!!!!!!!!!!

Answer 1

$\frac{8}{x+3} +\frac{6}{x-3} =1\frac{1}{5} \\\\\frac{8(x-3)}{(x+3)(x-3)} +\frac{6(x+3)}{(x-3)(x+3)} =\frac{6}{5} \\\\\frac{8x-24}{x^2-9} +\frac{6x+18}{x^2-9} =\frac{6}{5} \\\\\frac{8x-24+6x+18}{x^2-9} =\frac{6}{5} \\\\\frac{14x-6}{x^2-9} =\frac{6}{5}$

5(14x-6)=6(x²-9)

70x-30=6x²-54

6x²-54=70x-30

6x²-54-70x+30=0

6x²-70x-24=0

уравнение типа ax²+bx+c=0

D=b²-4ac=(-70)²-4×(-24)×6=5476

x=(-b±√D)/2a=(70±√5476)/12=(70±74)÷12=-1/3  и 12