Answer: ab =6
have:
[tex]\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}=1\\\\=>\frac{b}{ab}+\frac{a}{ab}+\frac{1}{ab}=1\\\\<=>\frac{a+b+1}{ab}=1[/tex]
=> a + b + 1 = ab
⇔ a + b + 1 - ab = 0
⇔ b - 1 - a(b - 1) + 2 = 0
⇔ (b - 1)(1 - a) = -2
because a and b are postive integers => (b - 1) and (1 - a) also are integers
=> (b - 1) ∈ {-1; 1; 2; -2;}
(1 -a) ∈ {-1; 1; 2; -2;}
because (b -1).(1-a) = -2 => we have the table:
b - 1 -1 1 2 -2
1 - a 2 -2 -1 1
a -1 3 2 0
b 0 2 3 -1
a.b 0 6 6 0
because a and b are postive integers
=> (a;b) = (3;2) or (a;b) = (2;3)
=> ab = 6
Step-by-step explanation:
