Question

若x+y=4,xy=3,则x^2+y^2= 若x^2 -2x+y^2 +6x+10=0,则x-y= 若m^2 +m-1=0,则m^3 +2m^2 +2011=

Answer 1

x^2+y^2
=(x+y)^2-2xy
=16-6
=10

x^2 -2x+y^2 +6y+10=0
(x-1)^2+(y+3)^2=0
x-1=0
y+3=0
x=1
y=-3
x-y=1+3=4

m^2 +m-1=0,
m^2 +m=1
m^3 +2m^2 +2011
=m（m^2+m+m）+2011
=m(1+m)+2011
=m^2+m+2011
=1+2011
=2012

Answer 2

解1：
x^2+y^2
=(x+y)^2-2xy
=4^2-2×3
=10

解2：
x^2-2x+y^2+6x+10=0
x^2-2x+1+y^2+6x+9=0
(x-1)^2+(y+3)^2=0
解得：x=1、y=-3
x-y=1-(-3)=4

解3：
m^2+m-1=0
m^2+m=1
m^3+2m^2+2011
=m(m^2+2m)+2011
=m[(m^2+m)+m]+2011
=m(1+m)+2011
=m^2+m+2011
=1+2011
=2012

Answer 3

第一个等于6 第二个等于4 第三个等于有点复杂口算不出来，