Question

(π-3)的0次方-|1-tan60°|+(-1/3)平方-根号5

Answer 1

(π-3)的0次方-|1-tan60°|+(-1/3)平方-根号5
=0 - (根号3-1)+(1/9)-根号5
=根号3-根号5+1/9

Answer 2

你的问题是：(π-3)的0次方-|1-tan60°|+(-1/3)平方-根号5
(π-3)^0-|1-tan60°|+(-1/3)²-√5
解：原式=1-（√3-1）+1/9-√5
=1+1-√3+1/9-√5
=19/9-√3-√5