问题
填空题
已知2x+y=1,代数式(y+1)2-(y2-4x)的值为______.
答案
∵2x+y=1,
∴(y+1)2-(y2-4x)
=y2+2y+1-y2+4x
=2y+4x+1
=2(2x+y)+1
=2×1+1
=2+1
=3.
故答案为:3.
已知2x+y=1,代数式(y+1)2-(y2-4x)的值为______.
∵2x+y=1,
∴(y+1)2-(y2-4x)
=y2+2y+1-y2+4x
=2y+4x+1
=2(2x+y)+1
=2×1+1
=2+1
=3.
故答案为:3.