解：?$(1)$?∵?$\sqrt 9＜\sqrt {11}＜\sqrt {16}，$?即?$3 ＜ \sqrt {11}＜ 4$?
∴?$\sqrt {11}$?的整數(shù)部分是?$3，$?小數(shù)部分是?$\sqrt {11}-3$?
?$ (2)$?∵?$3 ＜ \sqrt {11}＜ 4，$?∴?$-4＜-\sqrt {11}＜-3$?
則?$11 - 4＜11-\sqrt {11}＜11 - 3，$?即?$7＜11-\sqrt {11}＜8$?
∴?$11-\sqrt {11}$?的整數(shù)部分是?$7，$?小數(shù)部分?$m = 11-\sqrt {11}-7 = 4-\sqrt {11}$?
又∵?$3 ＜ \sqrt {11}＜ 4$?
∴?$11 + 3＜11+\sqrt {11}＜11 + 4，$?即?$14＜11+\sqrt {11}＜15$?
∴?$11+\sqrt {11}$?的整數(shù)部分是?$14，$?小數(shù)部分?$n = 11+\sqrt {11}-14=\sqrt {11}-3$?
∴?$m + n=(4-\sqrt {11})+(\sqrt {11}-3)=4-\sqrt {11}+\sqrt {11}-3 = 1$?
∵?$1$?是有理數(shù)，∴?$m + n$?是有理數(shù)
