Maths Question ( Laws of Indices ) thx =]

1. (3m4n5)2
= 32m4(2)n5(2)
= 9m8n10

2. (-x2)3(-3x2y)(-2x3y)0
= (-1)3x2(3)(-3)x2y(1)
= 3x6+2y
= 3x8y

3. (p2/q3)3 X (2q3/pq2)2 / (-4p2/q)3
= 22/(-4)3 p2(3)-2-2(3)q3(3)+(3-2)(2)-3
= -1/16 p-2q8
= -q8/16p2

4. (5y2x-3)(5-2x2y-1)/(-2x-2y4)
= 51+(-2)/(-2) x-3+2-(-2)y2+(-1)-4
= -1/10 xy-3
= -x/10y3

5. (3xy0)(9x-1y-2)0/27x-5y2
= 3x(1)/27x-5y2
= 1/9 x1-(-5)y-2
= x6/9y2

1. (3m^4 n^5 )^2
=9m^4*2 n^5*2
=9m^8 n^10

2. (-x^2)^3 (-3x^2 y) (-2x^3 y)^0
=(-x^2*3) (-3x^2 y) (1) By a^0=1
=3x^6+2 y
=3x^8 y

3. (p^2 / q^3 )^3 * (2q^3 / pq^2 ) ^2 / (-4p^2 / q)^3
=(p^2*3 / q^3*3) * (4q^3*2 / p^2q^2*2) / (-64p^2*3 / q^3)
=(p^6 * 4q^6 * q^3) / (q^9 * p^2q^4 * -64p^6)
= -1/16 * p^6-2-6 * q^6+3-9-4
= -1/16 p^ -2 q^ -4

4. [(5y^2 x^-3) (5^-2 x^2 y^-1)] / -2x^-2 y^4
=(5^-1 x^-3+2 y^2-1) / -2x^-2 y^4
= -1/10 x^-1+2 y^1-4
= -1/10 x y^-3

5. [(3xy^0)(9x^-1 y^-2)^0] / 27x^-5 y^2
=(3x * 1 * 1) / 27x^-5 y^2 By a^0=1
=1/9 x^1+5 y^-2
=1/9 x^6 y^-2

2007-09-14 19:18:37 補充：
更正(3.)3. (p^2 / q^3 )^3 * (2q^3 / pq^2 ) ^2 / (-4p^2 / q)^3=(p^2*3 / q^3*3) * (4q^3*2 / p^2q^2*2) / (-64p^2*3 / q^3)=(p^6 * 4q^6 * q^3) / (q^6 * p^2q^4 * -64p^6)= -1/16 * p^6-2-6 * q^6 3-6-4= -1/16 p^ -2 q^ -1

1. (3m^4 n^5 )^2
= 9m^8 n^10 #

2. (-x^2)^3 (-3x^2 y) (-2x^3 y)^0
= -x^6 (-3x^2 y) ( 1 )
= 3x^8 y #

3. (p^2 / q^3 )^3 * (2q^3 / pq^2 ) ^2 / (-4p^2 / q)^3
= ( p^6 / q^9 )* ( 4q^6 / p^2 q^4 ) / ( -64p^6 / q^3 )
= ( p^6 / q^9 )* ( 4q^6 / p^2 q^4 )* ( q^3 / -64p^6 )
= 1 / ( -16q^4 p^2 ) #

4. [(5y^2 x^-3) (5^-2 x^2 y^-1)] / -2x^-2 y^4
= 5^-1 x^-1 / -2x^-2 y^4
= x / -10y^4 #

5. [(3xy^0)(9x^-1 y^-2)^0] / 27x^-5 y^2
= 3x^6 / 27y^2
= x^6 / 9y^2 #

2007-09-14 19:10:23 補充：
number 4 is wrong.Audrey Hepburn and pan are right !!

2007-09-14 19:13:23 補充：
第3條就唔知邊個對=.=''

2007-09-15 20:59:08 補充：
椅? pan係未抄我ge第３條咁我姐係岩啦

1.(3m^4 n^5 )^2 (space係唔係乘)
=9m^8 n^10
2.(-x^2)^3 (-3x^2 y) (-2x^3 y)^0 (任何野^0=1)
=-x^6 (-3x^2 y)
=6x^8 y
3.(p^2 / q^3 )^3 * (2q^3 / pq^2 ) ^2 / (-4p^2 / q)^3
=(p^6 / q^9) * (4q^6 / p^2 q^4 ) / (-64p^6 /q^6 )
=(p^6 / q^9) * (4q^6 / p^2 q^4 ) * (q^6 / -64p^6 )
=(4p^6 q^12) / (-64p^8 q^13 )
=-1/(4p^2 q)
4.[(5y^2 x^-3) (5^-2 x^2 y^-1)] / -2x^-2 y^4
=[(5y^2 x^-3) (25^-1 x^2 y^-1)] / -2x^-2 y^4
=[1/5 y x^-1 ] / -2x^-2 y^4
=-x/(10 y^3)
5.[(3xy^0)(9x^-1 y^-2)^0] / 27x^-5 y^2
=(3x)/ (27x^-5 y^2)
=x^6 / 9y^2

2007-09-14 19:57:21 補充：
計錯少少野2. (-x^2)^3 (-3x^2 y) (-2x^3 y)^0 = -x^6 (-3x^2 y) ( 1 ) = 3x^8y3. (p^2 / q^3 )^3 * (2q^3 / pq^2 ) ^2 / (-4p^2 / q)^3 = ( p^6 / q^9 )* ( 4q^6 / p^2 q^4 ) / ( -64p^6 / q^3 ) = ( p^6 / q^9 )* ( 4q^6 / p^2 q^4 )* ( q^3 / -64p^6 ) = 1 / ( -16p^2 q^4)