a) Ta có (x - 3)2 > 0
(y - 2)2 > 0
3 > 0
=> (x - 3)2 + (y - 2)2 + 3 > 3
=> min (x - 3)2 + (y - 2)2 + 3 = 3 <=> x = 3; y = 2
b) Ta có |x + 50| > 0
(x - y)2 > 0
100 > 0
=> |x + 50| + (x - y)2 + 100 > 100
=> min |x + 50| + (x - y)2 + 100 = 100 <=> x = y = - 50