Question

计算(x-2)(x+2)-(x+1)(x-3)的值

计算
(1)(x_2)(x+2)-(x+1)(x-3)
(2)(2x-y)^2-4(x-y)(x+2y)
求两个代数式的值，要用平方差公式和完全平方公式解题，要过程
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Answer 1

1、
原式=x^2-4-[(x-1)+2][(x-1)-2]
=x^2-4-[(x-1)^2-4]
=x^2-4-(x^2-2x+1-4)
=x^2-4-x^2+2x-1+4
=2x-1

2、
原式=4x^2-4xy+y^2-4(x^2+2xy-xy-2y^2)
=4x^2-4xy+y^2-4x^2+4xy-8y^2
=-7y^2

Answer 2

(x_2)(x+2)-(x+1)(x-3) =2x-1

(2x-y)^2-4(x-y)(x+2y) =9y^2-8xy

Answer 3

(1)(x-2)(x+2)-(x+1)(x-3)
=(x^2-2^2)-(x^2+x-3x-3)
=2x-1
(2)(2x-y)^2-4(x-y)(x+2y)
=(4x^2-4xy+y^2)-4(x^2+2xy-xy-2y^2)
=9y^2-8xy

Answer 4

1.x^2-4-x^2+2x+3=2x-1
2.4x^2-4xy+y^2-4x^2+8y^2-4xy=9y^2-8xy