Question

初二计算题

1》[（4x-3y）／（3xy²z）]-[（2y-5z）／（3xy²z）]－[（4x＋5z）／（3xy²z）]
2》[3／（x-4）]－[24／（x²-16）]
3》[（a²＋b²）／（a－b）]－a－b
4》[7／（2x²+6x）]－[（11－2x²）／（9-x²）]＋2

Answer 1

1.原式=[(4x-3y-2y-5z-4x+5z)]/3xy2z
=(-5y)/3xy2
=-5/3xy
2.原式=3/(x-4) -24/(x2-16)
=(3x+12-24)/(x2-16)
=(3x-12)/[(x-4)·(x+4)]
=3/(x+4)
3.原式=[（a2+b2)/(a-b) ]-[(a+b)(a-b)/(a-b)]
=(a2-b2)/(a-b)
=2b2/(a-b)
4.原式=[7/2x(x+3)]-[(11-2x2)/(9-x2)]+2
=[7(x-3)/2x(9-x2)]-[(22x-4x3)/2x(9-x2)]+[(36x-4x3)/2x(9-x2)]
=(21x-21)/(18x-2x3)
注：平方是2，立方是3 上面没有乘号勿把x看做乘号