Răspuns :
/ z / = 0 <=> z = 0;
In cazul tau, |(x-2)(4-y)| =0 <=> ( x - 2 ) * ( 4 - y ) = 0 <=> x - 2 = 0 sau 4 - y = 0 <=>
x = 2; y = 4.
Bafta!
In cazul tau, |(x-2)(4-y)| =0 <=> ( x - 2 ) * ( 4 - y ) = 0 <=> x - 2 = 0 sau 4 - y = 0 <=>
x = 2; y = 4.
Bafta!
Exemlul tau nu e inspirat deoarece este un caz particular.
Daca luam exemplul:
|(x - 2)(4 - y)| = 6 atunci:
(x - 2)(4 - y) = ± 6
Divizorii lui 6 sunt 1; 2; 3; 6
Ecuatia 1:
(x - 2)(4 - y) = 6
a)
x - 2 = 2 => x = 4
4 - y = 3 => y = 1
b)
x - 2 = 1 => x = 3
4 - y = 6 => y = -2
c)
x - 2 = -2 => x = 0
4 - y = -3 => y = 7
d)
x - 2 = -1 => x = 1
4 - y = -6 => y = 10
s.a.m.d.
Ecuatia 2:
(x - 2)(4 - y) = -6
a)
x - 2 = -2 => x = 0
4 - y = 3 => y = 1
b)
x - 2 = -1 => x = 1
4 - y = 6 => y = -2
c)
x - 2 = 2 => x = 4
4 - y = -3 => y = 7
d)
x - 2 = 1 => x = 3
4 - y = -6 => y = 10
s.a.m.d.