方程两边同时进行通分可得
[5(x2-16)+96]/(x2-16)=[(2x-1)(x-4)+(3x-1)(x+4)]/(x2-16)不好意思 这是两道题这好似第一个 原式=(x+y)[(x-y)²+xy]=(x²-y²)(x-y)+x²y+xy²=x³-x²y-xy²+y³+x²y+xy²（-x²y-xy²+x²y+xy²抵消掉）=x³+y³你这是什么？？？？？？？？？第一个：(x+y)[(x-y)²+xy](x+y)[(x-y)²+xy]=(x²-y²)(x-y)+x²y+xy²=x³-x²y-xy²+y³+x²y+xy²（-x²y-xy²+x²y+xy²抵消掉）=x³+y³这是第二个：3（2x-1）(x+6)-5(x-3)(x+6) 3(2x-1)(x+6)-5(x-3)(x+6) =3(2x''+11x-6)-5(x''+3x-18) =6x''+33x-18-5x''-15x+90 =x''+18x+72 (注:''为平方)