(x+y)^n

(x^n+y^n)(x^n-y^n)-(x^n+y^n)²计算.(xⁿ+yⁿ)(xⁿ-yⁿ)-(xⁿ+yⁿ)²=(xⁿ+yⁿ

m(x-y)^n+n(y-x)^2n因式分解m(x-y)^n+n(y-x)^2n=m(x-y)^n+n(x-y)^2n=(x-y)^n（m＋n(x-y)^n）原式=m(x-y)^n+n(x-y)^2n=(x-y)^n［m+n(x-y)^n］

m(x-y)-n(y-x)因式分解m(x-y)-n(y-x)=m(x-y)+n(x-y)=(m+n)(x-y)不懂可以追问(x-y)(m+n)正确

3n【x+y】-x-y因式分解3n【x+y】-x-y=3n【x+y】-(x+y)=(3n-1)(x+y)如果本题有什么不明白可以追问,原式=3n(x+y)-(x+y)=(x+y)(3n-1)

m(x-y)+n(y-x)m(x?y)+n(y?x)=m(x?y)?n(x?y)=(m?n)(x?y)

x^n*y*(-x^2y^m)x^n*y*(-x^2y^m)=-x^(n+2)y^(1+m)

分解因式x^n+2*y^n-2+x^n+1*y^n-1+x^n+1*y^n-1x^(n+2)*y^(n-2)+x^(n+1)*y^(n-1)+x^(n+1)*y^(n-1)=x^(n+2)*y^(n-2)+2x^(n+1)*y^(n-1)=

x^2n-y^2n(因式分解)(x^n+y^n)(x^n-y^n)n(x+y)(x-y)(X^n+Y^n)(X^n-Y^n)=(X^n+Y^n)(X-Y)[X^(n-1)+X^(n-2)Y+X^(n-3)Y^2……+1]如果n为偶数，则(X

x^n-y^n的因式分解急用x^n-y^n=(x-y)[x^(n-1)+x^(n-2)y+x^(n-3)y^2+...+x^2y^(n-3)+xy^(n-2)+y^(n-1)]symsxyn>>factor(x^n-y^n)ans=x^n-

如何因式分解x^n+y^nx^n+y^n当n是奇数时=(x+y)[x^(n-1)-x^(n-2)y+x^(n-3)y^2-……-xy^(n-2)+y^(n-1)]n是偶数时不一能分解

因式分解（m+n)(x-y)-（m+n)(x+y)(m+n)(x-y)-(m+n)(x+y)=(m+n)[(x-y)-(x+y)]=(m+n)(-2y)=-2y(m+n)很高兴为您解答,【the1900】团队为您答题.请点击下面的【选为满意

(m-n)(x-y)-(n-m)(x+y)因式分解（m-n)(X-Y)-(n-m)(X+Y)=(m-n)(X-Y)+(m-n)(X+Y)=(m-n)(X-Y+X+Y)=2X(M-n)(m-n)(x-y)-(n-m)(x+y)=(m-n)(x

(m+n)(x-y)+(m-n)(y-x)解(m+n)(x-y)+(m-n)(y-x)=(m+n)(x-y)-(m-n)(x-y)=(x-y)(m+n-m+n)=2n(x-y)(m+n)(x-y)+(m-n)(y-x)=(x-y)（m+n-

4(x-y)^n+1-3(y-x)^n4(x-y)^n+1-3(y-x)^nn为奇数=(x-y)^n×[4(x-y)+3]=(x-y)^n×(4x-4y+3)n为偶数=(x-y)^n×[4(x-y)-3]=(x-y)^n×(4x-4y-3)

(m+n)(x-y)-(m+n)(x+y)的答案(m+n)(x-y)-(m+n)(x+y)=(m+n)[(x-y)-(x+y)]+(m+n)(-2y)=-2y(m+n)(因式分解答案）=-2my--2ny(计算题答案）

(m+n)(x-y)-(m+n)(x+y)的结果-2my-2ny-2my-2ny(m+n)(x-y)-(m+n)(x+y)=(m+n)[(x-y)-(x+y)]=(m+n)(x-y-x-y)=-2y(m+n)=-2y（x+y）=-2y(m+

(x+y)(m+n)2-(x+y)(m+n)2+Y=M+N反应前后质量保持不变,所以5gX和3gY完全反应,除生成1gM,其余的都是N,即生成N7g要制取14gN,即要两倍以上质量的反应物反应.即要10gX和6gY反应.根据质量首恒定律可知

计算题：(x+y)^n×(x+y)^n+1解(x+y)^n×(x+y)^n+1=(x+y)^(n+n+1)=(x+y)^(2n+1)(x+y)^n×(x+y)^(n+1)=(x+y)^[n+(n+1)]=(x+y)^(2n+1)

计算(x-y-m+n)(x-y+m-n)可以变成（x-y）-(m-n)乘（x-y）+(m-n)得到（x-y）的平方减（m-n）的平方

m(x+y)+n(x+y)-x-y因式分解m(x+y)+n(x+y)-x-y=m(x+y)+n(x+y)-(x+y)=(x+y)(m+n-1)m(x+y)+n(x+y)-x-y=(x+y)(m+n)-(x+y)=(x+y)(m+n-1)原式