已知a^2+2ab+b^2+(a+2)^2=0,b=计算:(x^2n-2x^ny^n+y^2n)/(x^n-y^n)=已知x+y=a,xy=b,则xy^2+yx^2=

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已知a^2+2ab+b^2+(a+2)^2=0,b=计算:(x^2n-2x^ny^n+y^2n)/(x^n-y^n)=已知x+y=a,xy=b,则xy^2+yx^2=已知a^2+2ab+b^2+(a+

已知a^2+2ab+b^2+(a+2)^2=0,b=计算:(x^2n-2x^ny^n+y^2n)/(x^n-y^n)=已知x+y=a,xy=b,则xy^2+yx^2=
已知a^2+2ab+b^2+(a+2)^2=0,b=
计算:(x^2n-2x^ny^n+y^2n)/(x^n-y^n)=
已知x+y=a,xy=b,则xy^2+yx^2=

已知a^2+2ab+b^2+(a+2)^2=0,b=计算:(x^2n-2x^ny^n+y^2n)/(x^n-y^n)=已知x+y=a,xy=b,则xy^2+yx^2=
1.原式 => (a+b)^2 + (a+2)^2 = 0
因为任何实数的平方>=0,所以a+b = 0; a+2 = 0.
所以a=-2,b=2
2.原式 = [(x^n)^2-2x^ny^n + (y^n)^2]/(x^n-y^n)
= (x^n-y^n)^2/(x^n-y^n)
= x^n - y^n
3.xy^2+yx^2 = xy(y+x) = ba = ab

第三道:=2xy*(x+y)=2b*a=2ab

已知a^2+2ab+b^2+(a+2)^2=0, b=
(a+b)^2+(a+2)^2=0
a+b=0 a+2=0
b=2
xy^2+yx^2=(x+y)xy=ab
第二题没看懂