用两种方法计算
展开全部
(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
=[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
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那个我看不懂那些符号你能不能写在纸上照下来。
追答
[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
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