Megoldás az x-re (a + b-x) / c + (a + c-x) / b + (c + b-x) / a + (4x) / (a + b + c) = 1-re?

# (A + b-x) / c + (a + c-x) / b + (c + b-x) / a + (4x) / (a + b + c) = 1 #

# => (A + bx) / c + 1 + (a + cx) / b + 1 + (c + bx) / a + 1 + (4x) / (a + b + c) -3-1 = 0 #

# => (A + b + c-x) / c + (a + c + b-x) / b + (c + b + a-x) / a-4 (1-x / (a + b + c)) = 0 #

# => (A + b + c-x) (1 / c + 1 / b + 1 / a) -4 ((a + b + c-x) / (a + b + c)) = 0 #

# => (A + b + c-x) (1 / c + 1 / b + 1 / a-4 / (a + b + c)) = 0 #

Így

# => (A + b + c-x) = 0 #

mert # (1 / c + 1 / b + 1 / a-4 / (a + b + c))! = 0 #

Ennélfogva # X = a + b + c #