Question

xy²/(x²y) - y×x²/(x²...

xy²/(x²y) - y×x²/(x²+x)= (x²-3x)/(x²-5x×2x) - 10/(x²-6x+9)= 2a/(a²-4) - 1/(a-2)= 2/(1-a ) - 1/(a-1)= 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

Answer 1

xy²/(x²y)
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y×x²/(x²+x)=y/x-x²y/x(x+1)=y/x-xy/x+1=((x+1)y-x²y)/x(x+1)=(xy+y-x²y)/x(x+1)
(x²-3x)/(x²-5x×2x)
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10/(x²-6x+9)=x(x-3)/(-9x²)-10/(x-3)²=(x-3)/(-9x)-10/(x-3)²=((x-3)^3+90x)/9x(3-x)
2a/(a²-4)
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1/(a-2)=2a/(a+2)(a-2)-1/(a-2)=(2a-a-2)/(a+2)(a-2)=1/(a+2)
2/(1-a
)
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1/(a-1)=(-2-1)/(a-1)=(-3)/(a-1)
x²-1/(x²-x-2)除以x/2x-4,其中x=1/2
解:原式=((x+1)(x-1)/(x+1)(x-2))/(x*2(x-2))
=(x-1)/(x-2)/x/2/(x-2)
=(x-1)/(2x(x-2)²)
当x=1/2时,原式=(-1/2)/(9/4)=-2/9
[x-x/(x+1)]除以[x/(2x-4)]
,其中x=√2+1
原式=((x(x+1)-x)/(x+1))/(x/(2(x-2)))
=(x²/(x+1))/(x/2/(x-2))
=x²/(x+1)/x*2*(x-2)
=2x(x-2)/(x+1)
当x=√2+1时,原式=(2√2+2)(√2+1)(√2-1)/(√2+2)
=(2√2+2)/(√2+2)
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