Algebra II help, simplifying

Simplify:
[ ( 2 - n )( 1/2 + 1/n ) ] / [ ( -4 + n^2 ) / ( 2n ) ]
The answer is apparently -1

Simplify:
[ ( 3x^2 + 12x +12 ) / ( 2x^2 + x - 6 ) ] * [ ( 4x^2 - 9 ) / ( 3x^3 + x^2 - 10x ) ]
The answer is apparently ( 6x + 9 ) / ( 3x^2 - 5x)

I have absolutely no idea what I did wrong on either of these.
I am supposed to correct both of these, but again, I am stuck. I have an answer key, but none of the work, which is what I am in need of.

Any help would be appreciated.

=[(2-n)(1/2+1/n)]/[(-4+n^2)/(2n)]
=[(2-n)(1/2+1/n)][(2n)]/[(-4+n^2)]
=[(2-n)(n+2)]/[(-4+n^2)]
=[(4-n^2)]/[(-4+n^2)]
=-[(-4+n^2)]/[(-4+n^2)]
=-1




=[(3x^2+12x+12)/(2x^2+x-6 )] *[(4x^2-9)/(3x^3+x^2-10x)]
=[(3x^2+12x+12)(4x^2-9)]/[(2x^2+x-6 )*(3x^3+x^2-10x)]
=3[x^2+4x+4][2x+3][2x-3]/[x*(3x^2+x-10… )]
=3[x+2][x+2][2x+3][2x-3]/[x*(3x^2+x-10…
=3[x+2][2x+3]/[x*(3x^2+x-10)]
=3[x+2][2x+3]/[x*(3x-5)(x+2)]
=3[2x+3]/[x*(3x-5)]
=(6x+9)/(3x^2-5x)

hope this helped u

[ ( 2 - n )( 1/2 + 1/n ) ] / [ ( -4 + n^2 ) / ( 2n ) ]=[(-1(n-2)(n+2)/2n]/[(n+2)(n-2)/2n]
=[-1(n-2)(n+2)2n]/[2n(n+2)(n-2)]
=-1

[ ( 3x^2 + 12x +12 ) / ( 2x^2 + x - 6 ) ] * [ ( 4x^2 - 9 ) / ( 3x^3 + x^2 - 10x ) ]=

[3(x²+4x+4)/{(2x-3)(x+2)}][(2x+3)(2x-3…‡
=( 6x+9)/[x(3x-5)]

PEMDAS, anything in parentheses, followed by exponents; ex: (3^2*6)=54 and so on.