（4/9)^(-sinπ/6)-2lg(cos0)+(0.125)^(log2 (1/3))=
问题描述：
（4/9)^(-sinπ/6)-2lg(cos0)+(0.125)^(log2 (1/3))=
答
(4/9)^(-sinπ/6)-2lg(cos0)+(0.125)^[log2 (1/3)]
=(4/9)^(-1/2)-2lg1+[2^(-3)]^[log2 (1/3)]
=(9/4)^(1/2)-0+2^[-3log2 (1/3)]
=3/2-0+2^[log2 (1/3)^(-3)]
=3/2+(1/3)^(-3)
=3/2+3³
=28.5
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答
（4/9)^(-sinπ/6)-2lg(cos0)+(0.125)^(log2 (1/8))
=（4/9)^(-1/2)-2lg1+(1/8)^(-3)
=(2/3)^(-1/2×2)-0+8³
=3/2+512
=513.5