Question

一道数学题我在线等

[(x^2-4)/x^2-4x+4 + 2-x/x+2]÷x/x-2 要没一步的过程谢谢

Answer 1

[(x^2-4)/(x^2-4x+4)+(2-x)/(x+2)]÷x/(x-2)
= [(x-2)(x+2)/(x-2)^2+(2-x)/x+2]÷x/(x-2)
= [(x+2)/(x-2)+(2-x)/x+2]÷x/(x-2)
= [(x+2)/(x-2)-(x-2)/x+2]÷x/(x-2)
= [(x+2)^2/(x-2)(x+2)-(x-2)^2/(x+2)(x-2)]÷x/(x-2)
=[(x+2)^2-(x-2)^2]/[(x+2)(x-2)]÷x/(x-2)
=8x/[(x+2)(x-2)]÷x/(x-2)
=8x/[(x+2)(x-2)]*(x-2)/x
=8/(x+2)

Answer 2

[(x^2-4)/x^2-4x+4 + 2-x/x+2]÷x/x-2
=[(x+2)(x-2)/(x+2)^2-(x-2)/(x+2)]×(x-2)/x
=[(x-2)/(x+2)-(x-2)/(x+2)]×(x-2)/x
=0×(x-2)/x
=0

Answer 3

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

Answer 4

(x^2-4)/x^2-4x+4 + 2-x/x+2÷x/x-2
=(x+2)(x-2)/(x-2)^2-2-x/2-x*x-2/x
=[(x+2)/(x-2)-1]*x-2/x
=(x+2)/(x-2)*(x-2)/x-(x-2)/x
=41/x