已知y+2x=1,求代数式(y+1)^2-(y^2-4x)

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已知y+2x=1,求代数式(y+1)^2-(y^2-4x)

已知y+2x=1,求代数式(y+1)^2-(y^2-4x)
已知y+2x=1,求代数式(y+1)^2-(y^2-4x)

已知y+2x=1,求代数式(y+1)^2-(y^2-4x)
原式=y^2+2y-1-y^2+4x
=2(y+2x)
=2×1
=2

(y+1)^2-(y^2-4x)
=y²+2y+1-y²+4x
=2y+4x+1
=2(y+2x)+1
=2+1
=3

因为y+2x=1
所以2x=1-y
所以
(y+1)²-(y²-4x)
=(y+1)²-(y²-2+2y)
=y²+2y+1-y²+2-2y
=3