解:∵?$1<\sqrt 2<2$?
∴?$3 + \sqrt 2$?的整數(shù)部分?$a = 4,$?小數(shù)部分?$b = 3 + \sqrt 2-4=\sqrt 2-1$?
?$3 - \sqrt 2$?的整數(shù)部分?$c = 1,$?小數(shù)部分?$d = 3 - \sqrt 2-1 = 2 - \sqrt 2$?
∴?$\frac {a - c}{b + d}=\frac {4 - 1}{\sqrt 2-1 + 2 - \sqrt 2}=\frac 31=3$?