Question

【急】求数学题详细解答过程

21lg20/3-lg7+21lg3+lg7/4
（lg5）^2+lg2*lg5+lg2

Answer 1

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

Answer 2

21 lg20/3 - lg7 + 21 lg3 + lg7/4
＝21 lg20－21 lg3－lg7 + 21 lg3 + lg7－lg4
＝21 lg10×2－lg2^2
＝21（lg10 + lg2）－2 lg2
＝21 + 21 lg2－2 lg2
＝21 + 19 lg2
(lg5)^2 + lg2 * lg5 + lg2
＝lg5 *（lg5 + lg2）+ lg2
＝lg5 * lg10 + lg2
＝lg5 + lg2＝lg10
＝1

Answer 3

都对