计算：(1)"1/6x-4y"+"1/6x+4y"-"3x/4y^2-9x^2" (2)1-(a- "1/1-a")^2除以"a^2-a+1/a^2-2a+1"

：(1)"1/6x-4y"+"1/6x+4y"-"3x/4y^2-9x^2"
=[(6x+4y)+(6x-4y)]/(6x+4y)(6x-4y)-3x/(2y+3x)(2y-3x)
=12x/4(3x-2y)(3x+2y)-3x/(2y+3x)(2y-3x)
=(12x+3x*4)/4(3x-2y)(3x+2y)
=6x/(9x^2-4y^2)
)1-(a- "1/1-a")^2除以"a^2-a+1/a^2-2a+1
=1-(a-a^2-1/1-a)^2/[a^2-a+1/(1-a)^2]
=(2a^2-3a+2)/(1-a)^2*[(1-a)^2/(a^2-a+1)]
=(2a^2-3a+2)/(a^2-a+1)

(1)"1/6x-4y"+"1/6x+4y"-"3x/4y^2-9x^2"
=(6x+4y+6x-4y)/[(6x+4y)(6x-4y)]-3x/[(2y-3x)(2y+3x)]
=12x/[(6x+4y)(6x-4y)]+3x/[(3x-2y)(3x+2yx)]
=3x/[(3x+2y)(3x-2y)]+3x/[(3x-2y)(3x+2yx)]
=6x/(9x^2-4y^2)
(2)1-(a- "1/1-a")^2除以"a^2-a+1/a^2-2a+1"
=1-[(a-a^2-1)/(1-a)]/[(a^2-a+1)/(a-1)^2]
=1+[(a^2-a+1)/(1-a)]/[(a^2-a+1)/(1-a)^2]
=1+[(a^2-a+1)(1-a)^2]/[(1-a)(a^2-a+1)]
=1+1-a
=2-a

1)"1/6x-4y"+"1/6x+4y"-"3x/4y^2-9x^2"
=1/(2(3x-2y))+1/(2(3x+2y))-3x/((2y-3x)(2y+3x))
=((3x+2y)+(3x-2y)+2*3x))/(2(3x-2y)(2y+3x))
=6x/((3x-2y)(2y+3x))
=6x/(9x^2-4y^2)
(2)1-(a- "1/1-a")^2除以"a^2-a+1/a^2-2a+1"
=[(1-a)^2+(1-a)]/(1-a)^2 除以"a^2-a+1/(a-1)^2"
=(2-3a+a^2)/(a^2-a+1)