三條數學問題

1. 2x^4*(-x^3)^2/(1/2x^2)^3
= 2x^4*x^6/(1/2^6x^6)
= 2x^10*2^3x^6
= 2^4x^16

2. (2/3a^2)^2*(ab^2)^3
= (2^2/3^2a^4)*a^3b^6
= (2^2*b^6)/(3^2*a)

3. (3a/2a-b)+(3b/2b-4a)
= (3a/2a-b)-(3b/4a-2b)
= (3a/2a-b)-(3b/2(2a-b))
= (6a-3b/2(2a-b))
= (3(2a-b))/(2(2a-b))
= 3/2

1)2x^4*(-x^3)^2/(1/2x^2)^3
=2x^4*(x^6)/[(1/2)^3(x^6)
=2x^4*1/(1/2)^3
=2^4x^4

2)(2/3a^2)^2*(ab^2)^3
=(2/3)^2a^4*a^3b^6
=(2/3)^2a^7b^6

3)(3a/2a-b)+(3b/2b-4a)
=3/2-b+3/2-4a
=3-4a-b

2007-12-14 20:03:16 補充：
is question 3 is 3a/(2a-b)+3b/(2b-4a)?If yes.=3a/(2a-b)-3b(4a-2b)=6a/(4a-2b)-3b/(4a-2b)=(6a-3b)/(4a-2b)=3(2a-b)/2(2a-b)=3/2