Question

1)(4b+1)(1-4b) 2)(a+3c)^2+(b+3c)(b-3c) 3)(x+3)^2-(x-3)^2 4)(x-4y)^2+(x+4y)^2 5)(x-3)(x++8)(x-8) 6)(p+1)^2-(p+2)^2 7)(y-2)(y--1)^2 8)3x(3x++1)^2 9)5b^2-(a-2b)^2 заранее огроменное человеческое

Answer 1

1)(4b+1)(1-4b) = 1 - 16b^2
2)(a+3c)^2+(b+3c)(b-3c) = a^2 + 6ac + 9c^2 + b^2 - 9c^2 = a^2 + 6ac + b^2
3)(x+3)^2-(x-3)^2 = (x+3-x+3)(x+3+x-3) = 12x
4)(x-4y)^2+(x+4y)^2 = x^2 - 8yx + 16y^2 + x^2 + 8yx + 16y^2 = 2x^2 + 32y^2
5)(x-3)(x+3)-(x+8)(x-8) = x^2 - 9 - x^2 +  64 = 55
6)(p+1)^2-(p+2)^2 = (p+1-p-2)(p+1+p-2) = 1-2p
7)(y-2)(y-3)-(y-1)^2 = y^2 - 5y + 6 - y^2 + 2y -1 = -3y + 5
8)3x(3x+7)-(3x+1)^2 = 9x^2 + 21x - 9x^2 - 6x - 1 = 15x - 1
9)5b^2-(a-2b)^2  = 5b^2 - a^2 + 4ab - 4b^2 = b^2 + 4ab - a^2

Answer 2

1) = (1+4b)(1-4b) = $1^{2} - (4b)^{2} = 1 - 16 b^{2}$
2) = $a^{2} +6ac + (3c)^{2} + b^{2} - (3c)^{2} = a^{2} +6ac+ + b^{2}$
3) = $x^{2} +6x + 9 x^{2} - ( x^{2} -6x+9 x^{2} ) = x^{2} +6x+9 x^{2} - x^{2} +6x-9 x^{2} =12x$
4) = $x^{2} -8xy+16 y^{2} + x^{2} +8xy+16 y^{2} = 2 x^{2} +32 y^{2}$
5) = $x^{2} -9-( x^{2} -64)= x^{2} -9- x^{2} +64=55$
6) = $p^{2} +2p+1-( p^{2} +4p+4)= p^{2} +2p+1- p^{2} -4p-4=-2p-3$
7) = $y^{2} -2y-3y+6-( y^{2} -2y+1)= y^{2} -5y+6- y^{2} +2y-1=-3y+5$
8) = $9x^{2} +21x-(9 x^{2} +6x+1)=9 x^{2} +21x-9 x^{2} -6x-1=15x-1$
9) = $5b^{2} -( a^{2} -4ab+4b^{2}) = 5 b^{2} - a^{2} +4ab-4b^{2} = b^{2} +4ab- a^{2}$