帮个忙 求你们了 急用xy²/(x²y-y) × x²/(x²+x)=(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=化简求植x²-1/(x²-x-2)除以x/2x-4,其中x=1/2[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1

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帮个忙 求你们了 急用xy²/(x²y-y) × x²/(x²+x)=(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=化简求植x²-1/(x²-x-2)除以x/2x-4,其中x=1/2[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1
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帮个忙 求你们了 急用xy²/(x²y-y) × x²/(x²+x)=(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=化简求植x²-1/(x²-x-2)除以x/2x-4,其中x=1/2[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1
帮个忙 求你们了 急用
xy²/(x²y-y) × x²/(x²+x)=
(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=
化简求植
x²-1/(x²-x-2)除以x/2x-4,其中x=1/2
[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1

帮个忙 求你们了 急用xy²/(x²y-y) × x²/(x²+x)=(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=化简求植x²-1/(x²-x-2)除以x/2x-4,其中x=1/2[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1
xy²/(x²y-y) × x²/(x²+x)=xy²/[(x+1)(x-1)y]×x²/[x(x+1)]=x²y/[(x+1)²(x-1)]
(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=x(x-3)/[x(x-5)]× 2(x-5)/(x-3)²=2/(x-3)
x²-1/(x²-x-2)=(x+1)(x-1)/[(x-2)(x+1)]=(x-1)/(x-2),将x=1/2代入,得(1/2-1)/(1/2-2)=1/3
[x-x/(x+1)]/[x/(2x-4)] =[x(x+1)-x]/(x+1)× [2(x-2)/x]=2x(x-2),将x=√2+1代入得
2*(√2+1)(√2+1-2)=2*(√2+1)(√2-1)=2