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如果x²%2xy+1+Axy+x²+xy%5合并同类项后,结果中...

整理后可得2*x2+(a-1)xy-4.若没有xy,所以,a-1=0.故a=1

-2+a+1=0a=1

=(x+y)(x-y)=[(x+y)(x-y)]=(x-y)=x^4-2xy+y^4

(x^2+2xy+y^2)(x^2-xy+y^2)^2=(x+y)^2*(x^2-xy+y^2)^2=[(x+y)(x^2-xy+y^2)]^2=(x^3+y^3)^2=x^6+2x^3y^3+y^6

此题中要运用x,y三次和的展开式,即 x^3+y^3=(x+y)(x^2-xy+y^2) ,知道这个式子后就好化简了原式=[(x+y)^2][(x^2-xy+y^2)^2] =(x^3+y^2)^2 由上面公式逆推即可 由于不知怎么表示幂次

如果,那么x+2xy+y=x+xy+(y+xy)=2+7=9如果a-ab=3,b+ab=2,那么a+b=(a-ab)+(b+ab)=3+2=5

1.原式为9,因为2个xy相加是2xy,x2+2xy+y2=2+7=92.原式为5,因为-ab与+ab抵消了,a2-ab+b2+ab=3+2=5

=(x+y)(x-xy+y)=[(x+y)(x-xy+y)]=(x+y)=x^6+2xy+y^6

x+y+2xy分之x-4y ÷x+xy分之2y+x=(x+2y)(x-2y)/(x+y)*x(x+y)/(x+2y)=x(x-2y)/(x+y)手机提问的朋友在客户端右上角评价点【满意】即可.

LZ的式子是(x/((x+2xy+y)/(x+y)))/(x-2xy+y)+(2x+2y)/(x-y)-2

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