Question

函数y=

log2(x-1)

|x-2|

的定义域为______．

Answer 1

根据题意，得

x-1＞0 |x-2|≠0

解此不等式组，得x＞1且x≠2
∴函数y=

log2(x-1)

|x-2|

的定义域为{x|x＞1且x≠2}
故答案为：{x|x＞1且x≠2}