问题 解答题
jm2y-[6my-2(umy-2)-m2y]+1,其中m=-
1
2
答案

3x1y-[6xy-1(4xy-1)-x1y]+1,

=3x1y-[6xy-8xy+4-x1y]+1,

=3x1y-6xy+8xy-4+x1y+1,

=4x1y+1xy-3,

当x=-

1
1
时,原式=4×(-
1
1
1y+1×(-
1
1
)y-3

=y-y-3

=-3.

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