Question

高一函数，给好评，有过程

Answer 1

1、解：原式=lg2(lg2+lg5)+lg5^2
=lg2+2lg5
=1+lg5
2、解：原式={(lg2)/(lg3)]+[lg2//(2lg3)]}{[lg3/(2lg2)]+[lg3/(3lg2)]}
=(3/2)(lg2/lg3)(5/6)(lg3/lg2)
=5/4
3、解：原式=lg5(3lg2+3)+3(lg2)^2-lg2-lg3+lg2+lg3-2
=3lg5lg2+3lg5+3(lg2)^2-2
=3lg2(lg5+lg2)+3lg5-2
=3lg2+3lg5-2
=3-2
=1
4、解：原式=10√3+10*[3^(1/4)]*[2^(-1/2)]*[3^(3/4)]*[2^(-1/2)]-10*(2+√3)
=10√3+(10*3)*2^(-1)-20-10√3
=15-20
=-5
5、解：原式=(3lg2+3lg5-lg2-lg5)/[(1/2)*(-1)]
=2(lg2+lg5)(-2)
=-4
6、解：原式=2lg3+lg7+2lg5-2lg3-lg3+2lg2
=2+lg7-lg3
7、解：原式=（25/9)(1/2)-[3^(-2)]-√(2-3)^2
=(25/18)-(1/9)-1
=5/18
8、解：原式=4+3-（1/16）
=111/16