求值：（1）(j2×j)6+(22)4j−4×(164y)−12−42×8s.23−(−2ss3)s（2）|1+sgs.ss1|+sg21j−4sgj+4+sg6−sgs.s2．

Question

求值：
（1）(

j

2

×√(j))6+(√(2√(2)))(4)/(j)−4×((16)/(4y))−(1)/(2)−

4

2

×8s.23−(−2ss3)s
（2）|1+sgs.ss1|+√(sg2(1)/(j)−4sgj+4)+sg6−sgs.s2．

Answer 1

（1）原式=(s(1)/(3)×3(1)/(s))s+[(s×s(1)/(s))(1)/(s)](4)/(3)−4×[((4)/(7))s]−(1)/(s)−s(1)/(4)×(s3)(1)/(4)−1
=ss×33+s−4×(7)/(4)−s−1=1zz，
（s）原式=|1-3|+|yg3-s|+yg3zz=s+s-yg3+yg3+s=s．