Question

已知m的平方+m-2=0,求m的三次方+3m的平方+2009的值？

Answer 1

m^2+m-2=0,m^2+m=2
m^2=2-m
m^3+3m^2+2009
=m^2(m+3)+2009
=(2-m)(m+3)+2009
=-m^2-m+6+2009
=-(m^2+m)+2015
=-2+2015
=2013,1,2013
由m^2+m-2=0 得m^2=2-m
代入入m^3+3m^2+2009,得
m(2-m)+3m^2+2009
化简后得2m^2+2m+2009
=2(m^2+m+2-2)+2009
=2(m^2+m-2)+4+2009
=0+4+2009
=2013
请采纳我的答案`,2,已知m的平方+m-2=0，
解这个一元二次方程得M=1或M=-2
当M=1时，m的三次方+3m的平方+2009的值=2013
当M=-2时，m的三次方+3m的平方+2009的值=2013,1,m^2 + m - 2 = 0
m^2 + m = 2
m^3 + 3m^2 + 2009 = (m^3 + m^2) + 2m^2 + 2009
= m(m^2 +m) + 2m^2 + 2009
= 2m + 2m^2 + 2009 = 2(m^2 + m)+ 2009
= 2*2 + 2009
= 2013,0,m^2=2-m,3m^2=6-3m,m的平方+m-2=0得到m^2+m+1=3,m^3-1=3m-3,m^3=3m-2,m^3+3m^2=6-3m+3m-2=4,答案2013,0,