（1）计算：0.25×(−12)−4−4÷(5−1)0−(116)−12；（2）计算：(169)−12+100(12lg9−lg2)+ln4e3+(log98)•(log433)．

Question

（1）计算：0.25×(−(1)/(2))−4−4÷(√(5)−1)0−((1)/(16))−(1)/(2)；
（2）计算：((16)/(9))−(1)/(2)+100((1)/(2)lg9−lg2)+ln

4

e3

+(log98)•(log4

3

3

)．

Answer 1

（1）原式=2−2×(2−1)−4−4÷1−(2−4)−(1)/(2)=2^-2+4-4-2²=4-4-4=-4；
（2）原式=[((3)/(4))−2]−(1)/(2)+102((1)/(2)lg32−lg2)+lne(3)/(4)+(lg8)/(lg9)×(lg3(1)/(3))/(lg4)=(3)/(4)+10lg((3)/(2))2+(3)/(4)+(3lg2)/(2lg3)×((1)/(3)lg3)/(2lg2)=(3)/(4)+(9)/(4)+(3)/(4)+(1)/(4)=(16)/(4)＝4．