F.1數學X2 {20分}

b^5 x (-12b)^2 ÷ (-3b^2)^3
=b^5 x (-12)^2 x b^2 ÷(-3)^3 x b^(2x3)
=b^5 x b^2 x 144 ÷ -27 x b^6
=b^(5x2) x 144 ÷ -27 b^6
=144b^10 ÷ -27b^6
=144÷-27 x b^(10-6)
= -16/3 b^4

(-6pq^2)^3 ÷ (2p^2q)^5 x (-8p^4q^6)^2
=(-60)^3 p^3 q^(2+3) ÷2^5 x p^(2+5) x q^5 x (-8)^2 x p^(4+2) x q^(6+2)
= -216000p^3q^5÷32p^7q^5 x -64p^6q^8
=(-216000÷32x-64) x p^(3-7+6) x q^(5-5+8)
=432000p^2q^8

自己再check多次睇下有無careless mistake同埋睇下明唔明啦

1....b^5 x (-12b)^2 ÷ (-3b^2)^3
=b^5(144b^2)/-27b^6
=144b^7/-27b^6
=16b/-3

1) b^5 x (-12b)^2 ÷ (-3b^2)^3
= b^5 x (-12)^2 x b^2 ÷ (-3)^3 ÷ b^6
= b^5 x 144 x b^2 ÷ (-27) ÷ b^6
= -144b/27
= -16b/3

2) (-6pq^2)^3 ÷ (2p^2q)^5 x (-8p^4q^6)^2
= (-6)^3 x p^3 x q^6 ÷ (2)^5 ÷ p^10 ÷ q^5 x (-8)^2 x p^8 x q^12
= -216 ÷ 32 x 64 x p^3 ÷ p^10 x p^8 x q^6 ÷ q^5 x q^12
= -432pq^13

1. (-240b)/(-18b)
=(6 2/3)b
2. (-18pq)*(-384pq)/20pq
=18pq*384pq/20pq
=345.6pq

1)b^5 x (-12b)^2 ÷ (-3b^2)^3
=b^5(144b^2)/-27b^6
=144b^7/-27b^6
=16b/-3

b^5 x (-12b)^2 ÷ (-3b^2)^3
=b^5x(144b^2)/(-27b^6)
=144b^7/-27^6
=-16b/3


(-6pq^2)^3 ÷ (2p^2q)^5 x (-8p^4q^6)^2
=(-6p^2q^2)^3/(2p^2q)^5x64p^8q^12
=p^(6-10+8)q^(6-5+12)*(-216)*64/32
=-432p^4q^13

2007-06-02 17:52:06 補充：
sor 第二題=p^(3-10+8)q^(6-5+12)......=-432pq^13 先啱

1. b^5 x (-12b)^2 / (-3b^2)^3
=b^5 x 144b^2 / -27b^8
=144b^7 /-27b^8
=5又1/3b

2. (-6pq^2)^3 / (2p^2q)^5 x (-8p^4q^6)^2
=-216p^3 q^8 / 32p^32 q^5 x 64p^8 q^12
=-512q^15 / p^21