1.已知1/x+1/y=5,则分式/的值为:

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1.已知1/x+1/y=5,则分式/的值为:2.已知x/=1/3,试求x^2/的值:

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2026-08-16 16:27:38

1/x+1/y=5[x+y]/xy=5x+y=5xy/=[2(x+y)-3xy]/[5xy+2xy]=[2*5xy-3xy]/[7xy]=7xy/7xy=1x/[x^2+x+1]=1/3[x^2+x+1]/x=3x+1/x+1=3x+1/x=2[x^4+x^2+1]/x^2=x^2+1/x^2+1=[x+1/x]^2-2+1=2*2-1=3所以:x^2/[x^4+x^2+1]=1/3

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  • 自由的曲奇
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    2026-08-16 16:27:38

    1/x+1/y=5[x+y]/xy=5x+y=5xy/=[2(x+y)-3xy]/[5xy+2xy]=[2*5xy-3xy]/[7xy]=7xy/7xy=1x/[x^2+x+1]=1/3[x^2+x+1]/x=3x+1/x+1=3x+1/x=2[x^4+x^2+1]/x^2=x^2+1/x^2+1=[x+1/x]^2-2+1=2*2-1=3所以:x^2/[x^4+x^2+1]=1/3

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