先化简后求值:(2x+y-1)(2x-y+1),求当x=2分之1,y=1时代数式的值

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/04 08:48:31
先化简后求值:(2x+y-1)(2x-y+1),求当x=2分之1,y=1时代数式的值

先化简后求值:(2x+y-1)(2x-y+1),求当x=2分之1,y=1时代数式的值
先化简后求值:(2x+y-1)(2x-y+1),求当x=2分之1,y=1时代数式的值

先化简后求值:(2x+y-1)(2x-y+1),求当x=2分之1,y=1时代数式的值
原式=[2x+(y-1)][2x-(y-1)]=4x^2-(y-1)^2
带入x=0.5,y=1得:原式=1

(2x+y-1)(2x-y+1)
=(2x)²-(y-1)²
=(2×1/2)²-(1-1)²
=1-0
=1


(2x+y-1)(2x-y+1)
=[2x+(y-1)][2x-(y-1)]
=(2x)²-(y-1)²
=(2×1/2)²-(1-1)²
=1²-0
=1

(2x+y-1)(2x-y+1)
=4x^2-(y-1)^2
当x=2分之1,y=1时
4x^2-(y-1)^2
=4*1/4-0
=1