a=2001²+2001²x2002²+2002²证明完全平方急
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![a=2001²+2001²x2002²+2002²证明完全平方急](/uploads/image/z/1162933-61-3.jpg?t=a%3D2001%26sup2%3B%2B2001%26sup2%3Bx2002%26sup2%3B%2B2002%26sup2%3B%E8%AF%81%E6%98%8E%E5%AE%8C%E5%85%A8%E5%B9%B3%E6%96%B9%E6%80%A5)
a=2001²+2001²x2002²+2002²证明完全平方急
a=2001²+2001²x2002²+2002²证明完全平方
急
a=2001²+2001²x2002²+2002²证明完全平方急
n=2001,n+1=2002
a=n^2+n^2*(n+1)^2+(n+1)^2
=n^2+n^2(n^2+2n+1)+(n^2+2n+1)
=n^2+n^4+2*n^3+n^2+n^2+2n+1
=n^4+2*n^3+3*n^2+2n+1
=n^2(n^2+n+1)+n(n^2+n+1)+(n^2+n+1) 这一步不太好明白,你可以试着倒回去算一下,是用拼凑的方法凑出来的.
=(n^2+2n+1)*(n^2+2n+1)
=(n^2+2n+1)^2完全平方数