一道数学化简题 3y^2+4y-7/(3y^3-3y)+(7y^2-7)

又不把括号括起来，谁看得懂

=3y^2-3+4y-4/3y(y^2-1)+7(y^2-1)
=3(y^2-1)+4(y-1)/(3y+7)(y^2-1)
=3(y+1)(y-1)+4(y-1)/(3y+7)(y^2-1)
=(3y+3+4)(y-1)/(3y+7)(y+1)(y-1)
=y-1/(y+1)(y-1)
=1/(y+1)

原式=（3y^2+4y-7）/[3y(y^2-1)+7(y^2-1)]
=（3y^2+4y-7）/(3y+7)(y^2-1)
=（3y^2+4y-7）/(3y+7)（y+1)(y-1)
=（3y^2+4y-7）/(3y^2+4y-7)(y+1) 注：3y+7与y-1相乘
=1/(y+1 )