Question

用两种方法计算 （3x/x-2 － x/x+2）* x的平方-4/x

Answer 1

解:(1)原式={[3x(x+2)-x(x-2)]/(x+2)(x-2)}*(x+2)(x-2)/x =3(x+2)-(x-2) =2x+8(2)原式=[3x/(x-2)]*(x+2)(x-2)/x-[x/(x+2)]*(x+2)(x-2)/x =3(x+2)-(x-2) =2x+8

Answer 2

（1）（3x/x-2 － x/x+2）* x的平方-4/x

=[3x（x+2)-x(x-2)](x^2-4)/(x^-4)x
=(3x^2+6x-x^2+2x)/x
=2x+8
(2)[3x(x^2-4)/(x-2) － x(x^2-4)/(x+2）]/x
=x[3(x^2-4)/(x-2) －x(x^2-4)/(x+2)]/x
=3(x+2)-(x-2)
=2x+8

Answer 3

[3x/(x-2)-x/(x+2) ]*(x²-4)/x 乘法分配律 (a+b)*c=ac+bc
=[3x/(x-2)]*(x²-4)/x-[x/(x+2) ]*(x²-4)/x
=3(x+2)-(x-2)
=2x+8

[3x/(x-2)-x(x+2) ]*(x²-4)/x 通分
={[3x(x+2)-x(x-2)]/(x+2)(x-2)}*(x²-4)/x
=[(3x²+6x-x²+2x)/(x²-4)]*(x²-4)/x
=2x+8

Answer 4

第一种，直接按顺序通分计算 第二种，先提公因式再算括号里的最后承开