作业帮(n+1)A^2-nB^2+AB=0怎样推算出(A+B)[(n+1)A-nB]=0
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(n+1)A^2-nB^2+AB=0怎样推算出(A+B)[(n+1)A-nB]=0
(n+1)A^2-nB^2+AB=0怎样推算出(A+B)[(n+1)A-nB]=0
(n+1)A^2-nB^2+AB=0怎样推算出(A+B)[(n+1)A-nB]=0
(n+1)A^2-nB^2+AB
=nA^2-nB^2+A^2+AB
=n(A^2 - B^2) + A(A+B)
=n(A+B)(A-B)+A(A+B)
=(A+B)[ n(A-B)+A ]
=(A+B)[(n+1)A-nB]
由(n+1)A^2-nB^2+AB=0得n（A^2-B^2)+A^2+AB再得n(A-B)(A+B)+A(A+B)再得(A+B)[(n+1)A-nB]=0