簡化代數式(中3)

[(-h^3 k²) ^-2] x ( h^6/k^-4 )

=[1 / (-h^3 k²)] ^2 x (h^6 x k^4)

例子：x^-1 = 1/x^1
[(-h^3 k^2) ^-2]..要先解決那-2次方，所以將其變成分數[1 / (-h^3 k^2)]^2
次方就會變成2　

( h^6 / k^-4 )的情況也一樣
將k^-4轉成分子，變成(h^6 x k^4)／1
即是=(h^6 x k^4), 次方就會變成4

=[1^2 / (-h^3)^2 (k^2)^2] x (h^6 x k^4)
例子：(x^2) ^2 = x^2x2
　　　　　　 = x^4
所以
(-h^3) ^2 = (-h) ^3x2
　　 =(-h)^ 6
=h^6

(k^2)^2 = k ^2x2
=k^4

=[1 / (h^6 k^4)] x (h^6 x k^4)

=(h^6 x k^4) / (h^6 x k^4)

=1 / 1

=1

[(-h^3 k^2) ^-2] x ( h^6/k^-4 )
=(h^-6 k^-4) x [ h^6/(1/k^4 )]
=(h^-6 k^-4) x ( h^6 k^4 )
=h^-6+6 k^-4+4
=h^0 k^0
=1

[(-h^3 k^2) ^-2] x ( h^6/k^-4 )
=1/[(-h^3 k^2) ^2] x [ h^6/(1/k^4 )]
=1/(h^6 k^4) x ( h^6 k^4 )
=1

[(-h^3 k^2) ^-2] x ( h^6/k^-4 )
re-arrange......

let a = [(-h^3 k^2) ^-2] & b = ( h^6/k^-4 ) for easier explaination

a = 1/ [(-h^3 k^2) ^2]
=> a = 1/ [(-h^3)^2 (k^2)^2]
=> a = 1/ [(-h)^ 6 k^4] because when we square negative numbers, the negative sign get cancel out
=> a = 1/ [h^ 6 k^4]

b = h^6 / k^4^(-1)
=> b = h^6 k^4 because k^4^-1 means 1/ (k^4)

combine a and b
a x b = 1/ [h^ 6 k^4] x h^6 k^4
and it cancels out, therefore, the solution is equal to ONE

[(-h^3 k^2) ^-2] x ( h^6/k^-4 )
=(h^-6 k^-4) x (h^6 k^4)
=1/h^6 k^4 x h^6 k^4
=h^6 k^4/h^6 k^4
=1

[(-h^3 k^2) ^-2] x ( h^6/k^-4 )
=(-h^3)^-2 x (k^2)^-2 x h^6 x k^4
=h^ -6 x k^ -4 x h^6 x k^4
=h^(-6+6) x k^(-4+4)
=h^0 x k^0
=1x1
=1