1.已知-3y=x+2z,求x^2-9y^2+4z^2+4xz的值.2.已知x不等于y,且x^2-x=13,y^2-y=13,求x+y-7的值.

1.已知-3y=x+2z,求x^2-9y^2+4z^2+4xz的值.
2.已知x不等于y,且x^2-x=13,y^2-y=13,求x+y-7的值.

1.已知-3y=x+2z,求x^2-9y^2+4z^2+4xz的值.
x^2-9y^2+4z^2+4xz
=(x+2z)^2-9y^2
=(-3y)^2-9y^2
=0
2.已知x不等于y,且x^2-x=13,y^2-y=13,求x+y-7的值.
x^2-x=13,y^2-y=13,两式相减的x^2-x－y^2+y=0
分解得,（x+y-1)(x-y)=0
因为 x≠y,即 x-y≠0,
所以 x+y-1＝0,
x＋y＝1
x+y-7＝1-7＝-6