Question

数学题1111

2(x+1)^2- 1/2(x+2)(x-2)
(2a+1)^2(2a-1)^2
(x-2)(x+2)-(x-1)(x-3)
(2a+1)^2-(2a^2+1)
(x+y+1)(x+y-1)
(2x-y+c)(-y-c-2x)

Answer 1

2(x+1)^2- (1/2)(x+2)(x-2)
=2(x^2+2x+1) - (1/2)(x^2-4)
=(3/2) x^2 + 4x

(2a+1)^2(2a-1)^2
= (4a^2+4a+1)(4a^2-4a+1)
=(4a^2+1 +4a)(4a^2+1 -4a)
=(4a^2+1)^2- (4a)^2
=16a^4+8a^2+1-16a^2
=16a^4-8a^2+1 =(4a^2-1)^2

(x-2)(x+2)-(x-1)(x-3)
=x^2-4 - (x^2-4x+3)
=x^2-4 - x^2+4x-3
=4x-7

(2a+1)^2-(2a^2+1)
=4a^2+4a+1- 2a^2-1
=2a^2+4a

(x+y+1)(x+y-1)
=(x+y)^2-1
=x^2+2xy+y^2-1

(2x-y+c)(-y-c-2x)
= (-1)(2x+c-y)(2x+c+y)
= - [(2x+c)^2-y^2]
= y^2 - (2x+c)^2
= y^2 - 4x^2 - 4xc - c^2

Answer 2

SF

追问

能算一下吗 50分