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【急】求数学题详细解答过程
21lg20/3-lg7+21lg3+lg7/4
(lg5)^2+lg2*lg5+lg2
人气:217 ℃ 时间:2019-11-20 04:48:51
解答
1.原式=21(lg20/3+lg3)- lg7 + lg7 - lg4
=21lg20-2lg2
=21(2lg2+lg5)-2lg2
=21(lg2+lg5)+21lg2-2lg2
=21+19lg2
2.原式=lg5(lg5+lg2)+lg2
=lg5+lg2
=1
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