f.3升f.4數學(指數)

1.)
(x^-1)(y^9) / (x^0)(y^7)
= y^9 / (x)(1)(y^7)
= y^2 / x


2.)
(^3√p^2)^6
= [p^(2/3)]^6
= p^4


3.)
9^4 x 9^-3 + 9^0
= 9^1 + 1
= 10


4.)
5^2 - (5^-13 / 5^-14)
= 5^2 - (5^14 / 5^-13)
= 5^2 - 5^1
= 25 - 5
= 20


5.)
121^(1/2)
= (11^2)^(1/2)
= 11


6.)
(13/11)^0 x (11/4)^-2
= 1 x (4/11)^2
= 16/121


7.)
5^(-3) - (1/5^3)
= (1/5^3) - (1/5^3)
= 0


8.) (2p^2/q)^5 / (-2p^2/-q^2)^3
= (2p^2/q)^5 / (2p^2/q^2)^3
= (2^5 p^10/q^5) / (2^3 p^6/q^6)
= 2^2 p^4/q^-1
= 4 p^4 q

2012-08-17 22:41:55 補充：
2.
我的理解是：
(³√p^2)^6
= [p^(2/3)]^6
= p^4

若題目是 (p^3√p^2)^6之誤，則：
(p^3√p^2)^6
= (p^3 x p)^6
= (p^4)^6
= p^24

1.)
(x^-1)(y^9)/(x^0)(y^7)
=y^2/x

2.)
(p^3√p^2)^6
=[(p^3)(p)]^6
=(p^4)^6
=p^24

3.)
9^4 X 9^-3 +9^0
=9+1
=10

4.)
5^2 - 5^-13 / 5^-14
=5^2-5
=5(5-1)
=5(4)
=20

5.)
121^1/2
=11

6.)
(13/11)^0 X (11/4)^-2
=(11^-2)/(4^-2)
=16/121

7.)
5^-3 -1/5^3
=(5^-3)-(5^-3)
=0

8.)
(2p^2/q)^5 / (-2p^2/-q^2)^3
=(32p^10/q^5)/(8p^6/q^6)
=(32p^10q^6)/(8p^6q^5)
=4p^4q