Question

已知x^2+xy=14,y^2+xy=2.求：..

Answer 1

1

x^2+xy=14

x(x+y)=14①

y^2+xy=2

y(x+y)=2②

①+②

x(x+y)+y(x+y)=16

(x+y)^2=16

x+y=±4

(x+y)^2-x^2+y^2

=(x+y)^2-(x+y)(x-y)

=(x+y)(x+y-x+y)

=2(x+y)y

由②

=2×2=4

2

x^2+y^2+4x-6y+4+9=0

(x^2+4x+4)+(y^2-6y+9)=0

(x+2)^2+(y-3)^2=0

所以x=-2,y=3

x^y

=(-2)^3=-8

3

(a/2)²=b

如a=4,则b=4

x^2+4x+4

=(x+2)^2

如a=8,b=16

x^2+8x+16

=(x+4)^2