初二几道分式题,a-1-(a²/a-1)(4/a+2)+a-2(1/x-3)-(1/x+3)
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初二几道分式题,a-1-(a²/a-1)(4/a+2)+a-2(1/x-3)-(1/x+3)
初二几道分式题,
a-1-(a²/a-1)
(4/a+2)+a-2
(1/x-3)-(1/x+3)
初二几道分式题,a-1-(a²/a-1)(4/a+2)+a-2(1/x-3)-(1/x+3)
a-1-(a²/a-1)=(a-1)*(a-1)/(a-1)-(a²/a-1)=(a-1)^2/(a-1)-(a²/a-1)=
(a^2-2a+1)/(a-1)-(a²/a-1)=(-2a+1)/(a-1)
(4/a+2)+a-2=(4/a+2)+(a-2)*(a+2)/(a+2)=(4/a+2)+(a^2-4)/(a+2)=a^2/(a+2...
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a-1-(a²/a-1)=(a-1)*(a-1)/(a-1)-(a²/a-1)=(a-1)^2/(a-1)-(a²/a-1)=
(a^2-2a+1)/(a-1)-(a²/a-1)=(-2a+1)/(a-1)
(4/a+2)+a-2=(4/a+2)+(a-2)*(a+2)/(a+2)=(4/a+2)+(a^2-4)/(a+2)=a^2/(a+2)
(1/x-3)-(1/x+3)=(1/x-3)*(x+3)/(x+3)-(1/x+3)*(x-3)/(x-3)=(x+3)/(x^2-9)-(x-3)/(x^2-9)=
6/(x^2-9)
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a-1-(a²/a-1)=[(a-1)^2-a^2]/(a-1)=(1-2a)/(a-1)
(4/a+2)+a-2=[4+(a-2)(a+2)]/(a+2)=a^2/(a+2)
(1/x-3)-(1/x+3)=[(x+3)-(x-3)]/[(x+3)(x-3)]=6/(x^2-9)