求助4m^2n^2-(m^2+n^2-x^2)^2因式分解!4m^2n^2-(m^2+n^2-x^2)^2 请问这道题的解答步骤应该是怎样的,希望高手前来指教~

原式=(2mn)^2-(m^2+n^2-x^2)^2
=-(m^2+n^2+2mn-x^2)(m^2+n^2-2mn-x^2)
=-[(m+n)^2-x^2][(m-n)^2-x^2]
=-(m+n+x)(m+n-x)(m-n+x)(m-n-x)

m^2-4n^2-4n-1
=m^2-(4n^2+4n+1)
=m^2-(2n+1)^2
=(m+2n+1)*(m-2n-1)

重复利用平方差公式
=(2mn+m^2+n^2-x^2)(2mn-m^2-n^2+x^2)
=[(m+n)^2-x^2][x^2-(m-n)^2]
=(m+n+x)(m+n-x)(x+m-n)(x-m+n)

4m^2n^2-(m^2+n^2-x^2)^2
=[2mn+(m^2+n^2-x^2)]*[2mn-(m^2+n^2-x^2)]
=[(2mn+m^2+n^2)-x^2]*[-(m^2-2mn+n^2)+x^2]
=[(m+n)^2-x^2]*[x^2-(m-n)^2]
=[(m+n+x)*(m+n-x)]*[(x+m-n)*(x-m+n)]
=(m+n+x)(m+n-x)](x+m-n)(x-m+n)

(2mn+m^2+n^2-x^2)(2mn-m^2-n^2+x^2)==(m+n+x)(m+n-x)(x-m+n)(x+m-n)

4m^2n^2-(m^2+n^2-x^2)^2
=(2mn)^2-(m^2+n^2-x^2)^2 化简
=(2mn+m^2+n^2-x^2)(2mn-m^2-n^2+x^2) 平方差
=[(m+n)^2-x^2][-(m-n)^2+x^2] 完全平方
=(m+n+x)(m+n-x)(x+m-n)(x-m+n) 平方差

4m^2n^2-(m^2+n^2-x^2)^2
=(2mn)^2-(m^2+n^2-x^2)^2
=(2mn+m^2+n^2-x^2)(2mn-m^2-n^2+x^2)
=[(2mn+m^2+n^2)-x^2](x^2-(m^2-2mn+n^2)]
=[(m+n)^2-x^2][x^2-(m-n)^2]
=(m+n+x)(m+n-x)(x+m-n)(x-m+n)

原式等于[2mn+(m^2+n^2-x^2)]*[2mn-(m^2+n^2-x^2)]
=[(2mn+m^2+n^2)-x^2]*[-(m^2-2mn+n^2)+x^2]
=[(m+n)^2-x^2]*[x^2-(m-n)^2]
=[(m+n+x)*(m+n-x)]*[(x+m-n)*(x-m+n)]
=(m+n+x)(m+n-x)](x+m-n)(x-m+n)

4m^2n^2-(m^2+n^2-x^2)^2
=[2mn+(m^2+n^2-x^2)]*[2mn-(m^2+n^2-x^2)]
=[(2mn+m^2+n^2)-x^2]*[-(m^2-2mn+n^2)+x^2]
=[(m+n)^2-x^2]*[x^2-(m-n)^2]
=[(m+n+x)*(m+n-x)]*[(x+m-n)*(x-m+n)]
=(m+n+x)(m+n-x)](x+m-n)(x-m+n)

原式=(2mn+m^2+n^2-x^2)(2mn-m^2-n^2+x^2)
=[(m+n)^2-x^2][x^2-(m-n)^2]
=(m+n+x)(m+n-x)(x+m-n)(x-m+n)

4m^2n^2-(m^2+n^2-x^2)^2
原式=(2mn)^2-(m^2+n^2-x^2)^2
=[2mn+(m^2+n^2-x^2)]*[2mn-(m^2+n^2-x^2)]
=[(2mn+m^2+n^2)-x^2]*[-(m^2-2mn+n^2)+x^2]
=[(m+n)^2-x^2]*[x^2-(m-n)^2]
=[(m+n+x)*(m+n-x)]*[(x+m-n)*(x-m+n)]
=(m+n+x)(m+n-x)](x+m-n)(x-m+n)