Anonymous

(b-4/b)/(4/b) = (b-4/b)(b/4) = (b^2)/4-1

Anonymous

(b - 4/b) ÷ (4/b) = b^2/4 - 1

Anonymous

(b-4/b)÷(4/b) In accord with the PEMDAS rules for priority and order of operations: = b/(4/b) - (4/b)/(4/b) = b²/4 - 1 = (b²-4)/4 : with the stipulation that b ≠ 0

Anonymous

(b - 4/b) / (4/b) = ((b^2 - 4) / b) / (4/b) = (b^2 - 4) / 4 for all nonzero b. (Or (b^2)/4 - 1.) Or: ((b - 4)/b) / (4/b) = (b - 4) / 4. (Or b/4 - 1. Again, this is for all nonzero b.)

Anonymous

you are getting various answers because the question is ambiguous case 1: (b-(4/b))÷(4/b) (b-(4/b)) / (4/b) multiply by b/b (b² – 4) / 4 or (b²/4) – 1 case 2: ((b-4)/b)÷(4/b) ((b-4)/b) / (4/b) multiply by b/b (b–4)/4 or (b/4) – 1

Anonymous

. (b - 4/b) ÷ 4/b = 1/b (b² - 4) × b/4 = ¼ ( b² - 4 ) = ¼ (b - 2)(b + 2) ————————

Anonymous

(b - 4/b) ÷ (4/b) = (b^2 - 4)/b ÷ (4/b) = (b^2 - 4) ÷ 4 = b^2/4 - 1

Anonymous

(b^2-4)/4

Anonymous

- 1 + ( b / 4 ) , b ╪ 0...assuming ( b - 4 ) / b divided by 4 / b