问题: 求助初二数学题
1.计算：1-[a-(1/1-a)]除以a^2-a+1/a^2-2a+1
2.当x=3时，求下列式子的值：(3x+2/x^2-x-2)+[1-(1/x+1)]除以[1+（1/x-1）]
3.先化简，再计算：(x^2+2x/1+x)除以[x-(2/x+1)],其中x=-1/2
4.若x/2=y/3,则分式7x^2-y^2/x^2-2xy+3y^2的值是多少？
解答:
(1)1-[a-1/(1-a)]÷[(a^-a+1)/(a^-2a+1)]
= 1-[(a-a^-1)/(1-a)]×(a-1)^/(a^-a+1)
= 1-[(a^-a+1)/(a-1)]×(a-1)^/(a^-a+1)
= 1-(a-1)
= 2-a
(2)(3x+2)/(x^-x-2)+[1-1/(x+1)]÷[1+1/(x-1)]
= (3x+2)/[(x+1)(x-2)]+[x/(x+1)]÷[x/(x-1)]
= (3x+2)/[(x+1)(x-2)]+(x-1)(x-2)/(x+1)(x-2)
= (3x+2+x^-3x+2)/[(x+1)(x-2)]
= (x^+4)/[(x+1)(x-2)]
= (9+4)/(4*1) = 13/4
(3)(x^+2x)/(1+x)÷[x-2/(x+1)].......其中x=-1/2
= (x^+2x)/(1+x)÷[(x^+x-2)/(x+1)]
= (x^+2x)/(x^+x-2)
= x(x+2)/[(x+2)(x-1)]
= x/(x-1) = (-1/2)/(-3/2) = 1/3
(4) 令x=2t，y=3t
（7x^-y^)/(x^-2xy+3y^) = (28-9)/(4-12+27) = 17/19
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