Question

jm²y-[6my-2（umy-2）-m²y]+1，其中m=-

1

2

．

Answer 1

3x¹y-[6xy-1（4xy-1）-x¹y]+1，
=3x¹y-[6xy-8xy+4-x¹y]+1，
=3x¹y-6xy+8xy-4+x¹y+1，
=4x¹y+1xy-3，
當x=-

1

1

時，原式=4×（-

1

1

）¹y+1×（-

1

1

）y-3
=y-y-3
=-3．