(log2)^3+(log5)^3+log5*log8

求解
2025-06-21 18:38:15
推荐回答(1个)
回答1:

令a=lg2,b=lg5
则lg2+lg5=lg10=1
lg8=lg2³=3lg2=3a
所以原式=a³+b³+b*3a
=(a+b)(a²-ab+b²)+3ab
=1*(a²-ab+b²)+3ab
=a²+2ab+b²
=(a+b)²
=1