问题
解答题
先化简,再求值.已知x=-1,y=2,求-(x2+y2)+[-3xy-(x2-y2)]的值.
答案
-(x2+y2)+[-3xy-(x2-y2)]
=-x2-y2+[-3xy-x2+y2]
=-x2-y2-3xy-x2+y2
=-2x2-3xy,
当x=-1,y=2时,
原式=-2x2-3xy
=-2×(-1)2-3×(-1)×2
=-2-(-6)
=4.
先化简,再求值.已知x=-1,y=2,求-(x2+y2)+[-3xy-(x2-y2)]的值.
-(x2+y2)+[-3xy-(x2-y2)]
=-x2-y2+[-3xy-x2+y2]
=-x2-y2-3xy-x2+y2
=-2x2-3xy,
当x=-1,y=2时,
原式=-2x2-3xy
=-2×(-1)2-3×(-1)×2
=-2-(-6)
=4.