先化简,再求值1/a-12/(a^2-1)-6/(1-a),其中a=1/2已知x/y=1/2,求(x^2+2xy+y^2)/(x^2-xy+y^2)的值

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先化简,再求值1/a-12/(a^2-1)-6/(1-a),其中a=1/2已知x/y=1/2,求(x^2+2xy+y^2)/(x^2-xy+y^2)的值
先化简,再求值1/a-12/(a^2-1)-6/(1-a),其中a=1/2
已知x/y=1/2,求(x^2+2xy+y^2)/(x^2-xy+y^2)的值

先化简,再求值1/a-12/(a^2-1)-6/(1-a),其中a=1/2已知x/y=1/2,求(x^2+2xy+y^2)/(x^2-xy+y^2)的值
1/a-12/(a^2-1)-6/(1-a)=1/a+12/(1-a^2)-6(1+a)/(1-a)(1+a)
=1/a+12/(1-a^2)-6(1+a)/(1-a^2)
=1/a+12-6(1+a)/(1-a^2)
因为1-a不为零,所以上式为
1/a+6/1+a
将a=1/2代入上式,2+6/1+0.5=6
问题补充答案
(x^2+2xy+y^2)/(x^2-xy+y^2),上下两式同除以y^2,有
[(x/y)^2+2x/y+1]/[(x/y)^2-x/y+1],将x/y=1/2代入上式
(1/4+1+1)/(1/4-1/2+1)=3
因为x/y=1/2,设x=k,y=2k
原式=(x+y)^2/(x^2-xy+y^2-xy+xy)=(x+y)^2/[(x-y)^2+xy]
将 x=k,y=2k 代入上式,则
(k+2k)^2/(k^2+2k^2)=3

(x²+2xy+y²)(x²-xy+y²)²
=(x+y)²(x²-xy+y²)²
=[(x+y)(x²-xy+y²)]²
=(x³+y³)²
=x^6+2x³y³+y^6
2.
a...

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(x²+2xy+y²)(x²-xy+y²)²
=(x+y)²(x²-xy+y²)²
=[(x+y)(x²-xy+y²)]²
=(x³+y³)²
=x^6+2x³y³+y^6
2.
a+b+c=0
a=-(b+c)
a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)
=-(b+c)(1/b+1/c)+b/c+c/b+(b+c)/[-(b+c)]
=-1-c/b-1-b/c+b/c+c/b-1
=-3

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