初一数学题两道（同底数幂）求详细步骤1 m²+m-1=0,求m³+2m²+20102 x²+x-2009=0,求x³-2010x+2010

1.因为m²+m-1=0，
所以m²+m=1。
m³+2m²+2010
=（m³+m²）+m²+2010
=m（m²+m）+m²+2010
=mx1+m²+2010
=m+m²+2010
=1+2010
=2011
2. 因为 x²+x-2009=0
所以x²+x=2009
x³-2010x+2010
=x³+x²-x²-2010x+2010
=x(x²+x)-x²-2010x+2010
=2009x-x²-2010x+2010
=-x²-x+2010
=-(x²+x)+2010
=-2009+2010
=1

1 m2+m-1=0,
m3+2m2+2010
=m3+m2-m+m2+m+1+2009
=m(m2+m-1)+m2+m+1+2009
=2009
2 x2+x-2009=0,
x3-2010x+2010
=x3+ x2- x2-2009x-x+2010
=x(x2+x-2009）-x2-x+2009+1
=1

一题 m³+2m²+2010
=m³+m²+m²+2010
=m(m²+m)+m²+2010 （m²+m=1)
=m+m²+2010 (同理）
=2011
二题x³-2010x+2010
=x³+x²-2009x-x²-x+2010
=x(x²+x-2009)-x²-x+2010 (x²+x-2009=0)
=-(x²+x)+2010 (x²+x=2009)
=-2009+2010
=1

1 m²+m-1=0
∴m²+m=1
m³+2m²+2010
=m(m²+m)+m²+2010
=m+m²+2010
=2011
2 x²+x-2009=0
∴ x²+x=2009
x³-2010x+2010
=x³+x²-x²-2010x+2010
=x(x²+x)-x²-2010x+2010
=2009x-x²-2010x+2010
=-x²-x+2010
=-2009+2010
=1

m²+m-1=0
∴
m³+2m²+2010
=m³+m²+m²+2010
=m(m²+m)+m²+2010
=m+m²+2010 [注：m²+m=1，则：m(m²+m)=m]
=1+2010
=2011
x²+x-2009=0
∴ x²+x=2009
x³-2010x+2010
=x³+x²-x²-2010x+2010
=x(x²+x)-x²-2010x+2010
=2009x-x²-2010x+2010 [注：x²+x=2009，则x(x²+x)=2009x]
=-x²-x+2010
=-(x²+x)+2010
=-2009+2010
=1

1, m(m^2+m+1)+m^2-m+2010=2011-2m=2011-(-1+-根号5）=2012+-根号5
2，x（x^2+x-2009)-(x^2+x)+2010=2009+2010=4019

m²+m=m(m+1)=1,
m³+2m²=m²(m+2),
. =mxm(m+m1+1),
=mx[m(m+1)+m],
∵m(m+1)=1，∴mxm[m(m+1)+m]=m(1+m),
=1+2010
=1

1.因为m²+m-1=0,所以m²+m=0+1=1,
m³+2m²+2010
=m·m²+m²+m²+2010
=m·m²+m·m+m²+2010
=m（m+m²）+m²+2010
=1·m+m²+2010
=m²+m+2010
=1+2010
=2011
2.因为x²+x-2009=0,所以x²+x=2009
x³-2010x+2010
=x³+x²-x²-2010x+2010
=x²x+x·x-x²-2010x+2010
=x（x²+x）-x²-2010x+2010
=2009x-x²-2010x+2010
=-x²-x+2010
=-（x²+x）+2010
=-2009+2010
=1

1）m^2+m-1=0 m^2+m=1
m^3+2m^2+2010=m(m^2+m)+m^2+2010= m+m^2+2010=2011
2)x^2+x-2009=0
x^3-2010x+2010=x(x^2+x-2009)-x^2-x+2010=-(x^2+x-2009)+1=1