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初二几道分式计算
1.b/(a-b)+a/(a+b)-2ab/(b^2-a^2)
2.(2a+2)/(a+1)-(a^2-1)/(a^2-2a+1)
3.(x^2-2x)/x^2-1/[x-1-(2x-1/x+1)
人气:463 ℃ 时间:2020-05-13 00:58:01
解答
1.=b(a+b)/(a-b)(a+b)+a(a-b)/(a+b)(a-b)-2ab/(b^2-a^2)
=(ab+b^2+a^2-ab)/(a^2-b^2)+2ab/b^2-a^2
=a^2+b^2/(a^2-b^2)-2ab/a^2-b^2
=(a-b)^2/(a+b)(a-b)
=a-b/a+b
2.=2(a+1)/(a+1)-(a^2-1^2)/(a-1)^2
=2-(a+1)(a-1)/(a-1)^2
=2-(a+1)/(a-1)
3 =x(x-2)/(x^2-1^2)-1/{(x+1)(x-1)/(x+1)++(2x-1)/(x+1)
=x(x-2)/(x^2-1^2)-1/(x^2-2x)/(x+1)
=x(x-2)/(x^2-1^2)-(x+1)/x(x-2)
=x+1/(x+1)(x-1)
=1/(x-1)
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