Log^x+1 (x-1/2) = log^x-1/2 (x+1)

условимся: рядом логарифмом пишу основание потом число

log (x+1) (x-1/2) = 1/log (x+1) (x-1/2)

log (x+1) (x-1/2) - 1/log (x+1) (x-1/2) = 0

(log² (x+1) (x-1/2) - 1=0. log (x+1) (x-1/2) ≠0

(log (x+1) (x-1/2) - 1) * (log (x+1) (x-1/2) + 1) = 0. x>1/2. x≠3/2

log (x+1) (x-1/2) - 1=0 или log (x+1) (x-1/2) + 1=0

log (x+1) (x-1/2) = 1 log (x+1) (x-1/2) = - 1

x+1=x-1/2 x-1/2=1 / (x+1)

нет решения (x-1/2) (x+1) - 1=0

x²-1/2x+x-1/2-1=0

x²+1/2x-3/2=0 / * 2

2x²+x-3=0 D=1²-4*2*-3=25

x1=-1+5) / 4=4/4=1 x2=-1-5) / 4=-6/4=-3/2

-3/2-постороний корень

ответ 1