初3的分式问题

来源：百度知道 编辑：UC知道 时间：2024/05/12 18:16:33
化简求值:(1)(x2+4/x2-4)-(2/x-2) x=(根号2)-2
(2)(x2-6x+9/2x-6)*(x+3) x=根号5
(3)(x/x-1)-(3/(x-1)(x+2))-1 并说出(3)中x取值范围!

(1)
(x2+4/x2-4)-(2/x-2)
=(x^2+4-2(x+2))/(x^2-4)
=(x^2-2x)/(x^-4)
=x(x-2)/(x-2)(x+2)
=x/(x+2)
=1-2/(x+2)
=1-2/(√2-2+2)
=1-2/√2
=1-√2

(2)
(x2-6x+9/2x-6)*(x+3)
=(x-3)^2/2(x-3)*(x+3)
=(x-3)(x+3)/2
=(x^2-9)/2
=((根号5)^2-9)/2
=(5-9)/2
=-2

(3)(x/x-1)-(3/(x-1)(x+2))-1
=(x(x+2)-3)/(x-1)(x+2)-1
=(x^2+2x-3)/(x-1)(x+2)-1
=(x-1)(x+3)/(x-1)(x+2)-1
=(x+3)/(x+2)-1
=1/(x+2)

x取值范围:x≠1且x≠-2

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