已知m平方+m-1=0 则m三次方+2m平方+1999

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已知m平方+m-1=0 则m三次方+2m平方+1999

已知m平方+m-1=0 则m三次方+2m平方+1999
已知m平方+m-1=0 则m三次方+2m平方+1999

已知m平方+m-1=0 则m三次方+2m平方+1999
m^3+2m^2+1999
=m(m^2+m)+m^2+1999
因为m^2+m-1=0
所以m^2+m=1
于是,
上式=m*1+m^2+199
=m^2+m+1999
=1+1999
=2000

m^2+m-1=0
m(m+1)=1
m^3+2m^2+1999
=m^2(m+1)+m^2+m-m-1+2000
=m^2(m+1)+m(m+1)-(m+1)+2000
=(m^2+m-1)(m+1)+2000
=2000

m^2+m-1=0
m(m+1)=1
m^3+2m^2+1999
=m^2(m+1)+m^2+m-m-1+2000
=m^2(m+1)+m(m+1)-(m+1)+2000
=(m^2+m-1)(m+1)+2000
=2000

M^2+M=1
原式=M^3+2M^2+1999=M^3+M^2+M^2+1999
=M(M^2+M)+M^2+1999
=M+M^2+1999
=1+1999
=2000