Question

这些数学题怎么做？

1:若a-b=2,ab-c=1,则(2a-b-c)^2+(c-a)^2=
2:如果x+y=-1,x-y=-3,那么x^2-y^2=
3:已知m^2+m-1=0,则m^3+2m^2+2004=

Answer 1

a=b+2
c=a-1=b+1
(2a-b-c)^2+(c-a)^2=3^2+1^2=10

x^2-y^2=（x+y）(x-y)=3

m^2+m-1=0
同乘m
m^3+m^2-m=0
原式就变成m^2+m+2004
=2005

Answer 2

1.第一个条件是不是错了
应该是a-b=2，a-c=1吧，不然好像没法解答
(2a-b-c)^2+(c-a)^2
=(a-b+a-c)^2+(a-c)^2
=(2+1)^2+(1)^2
=10

2.x^2-y^2
=(x+y)(x-y)
=-1×(-3)
=3

3.由m^2+m-1=0
得m^2+m=1
m^3+2m^2+2004
=m^3+m^2+m^2+2004
=m(m^2+m)+m^2+2004
=m+m^2+2004
=1+2004
=2005

解答完毕！