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log3 4√27/3+lg125/2+lg16-√1/16+(-√2)^-2
人气:321 ℃ 时间:2020-07-29 13:23:34
解答
log3 [(四次根号27)/3] + lg(125/2) + lg16 - 根号(1/16) + (-√2)^2 -2
= log3 (3^(-1/4)) + lg(5^3 /2) + lg(2^4) - 1/4 + 2 - 2
= -1/4 + 3lg 5 - lg 2 + 4lg 2 - 1/4
= -1/2+3lg 5 +3lg2
= -1/2+3lg (5*2)
= -1/2+3lg 10
= -1/2+3
= 5/2
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