求下列各式的值(1) ½lg(32/49)- 3/4lg(根号8) + lg(根号245)(2)(lg5)² + 2lg(2) - (lg2)²

问题描述:

求下列各式的值
(1) ½lg(32/49)- 3/4lg(根号8) + lg(根号245)
(2)(lg5)² + 2lg(2) - (lg2)²

(1)
1/2 lg(32/49)- 3/4lg(根号8) + lg(根号245)
=1/2(lg(2^5)-lg(7^2))- 3/4lg(2^(3/2)) + lg(7*5^(1/2))
=5/2 lg2 -lg7 -9/8 lg2 +lg7+ 1/2 lg5
=11/8 lg2 -1/2 lg2
=7/8 lg2.
(2)
lg5=lg(10/2)=lg10-lg2=1-lg2,
(lg5)² + 2lg(2) - (lg2)²
=(1-lg2)^2+ 2lg(2) - (lg2)^2
=1.

⑴1/2lg(32/49)-3/4lg(√8)+lg(√245)
=1/2(lg32-lg49)-3/4lg[2^(3/2)]+lg(7×√5)
=1/2(5lg2-2lg7)-3/4×3/2lg2+lg7+1/2lg5
=5/2lg2-lg7-9/8lg2+lg7+1/2lg5
=11/8lg2+1/2lg5
=7/8lg2+1/2lg2+1/2lg5
=7/8lg2+1/2(lg2+lg5)
=7/8lg2+1/2
此时怀疑是抄错题,
1/2lg(32/49)-4/3lg(√8)+lg(√245)
=1/2(lg32-lg49)-4/3lg[2^(3/2)]+lg(7×√5)
=1/2(5lg2-2lg7)-4/3×3/2lg2+lg7+1/2lg5
=5/2lg2-lg7-2lg2+lg7+1/2lg5
=1/2lg2+1/2lg5
=1/2(lg2+lg5)
=1/2
⑵(lg5)^2+2lg2-(lg2)^2
=(lg5)^2-(lg2)^2+2lg2
=(lg5+lg2)(lg5-lg2)+2lg2
=lg5-lg2+2lg2
=lg5+lg2
=1