1) { (4-y)^2 - y(y+1), y = -1/9 } 2) { (2-c)^2 - c(c+4), c = 0.5 } 3) { (8b-8)(8b+8) - 8b(8b+8), b = 2.6 } 4) { x^3+x-(x-2)(x^2+2x+4), x=0.0015 } 5) { (a-3)^2 - 6(2-a), a=0.5 } 6) { -p(4+p)+(p-2)(p+2) }

Решение:

1. \( (16 - 8y + y^2) - (y^2 + y) = 16 - 9y \). При \( y = -\frac{1}{9} \): \( 16 - 9(-\frac{1}{9}) = 16 + 1 = 17 \)
2. \( (4 - 4c + c^2) - (c^2 + 4c) = 4 - 8c \). При \( c = 0.5 \): \( 4 - 8(0.5) = 4 - 4 = 0 \)
3. \( (64b^2 - 64) - (64b^2 + 64b) = 64b^2 - 64 - 64b^2 - 64b = -64b - 64 \). При \( b = 2.6 \): \( -64(2.6) - 64 = -166.4 - 64 = -230.4 \)
4. \( x^3+x - (x^3 - 8) = x^3 + x - x^3 + 8 = x+8 \). При \( x=0.0015 \): \( 0.0015 + 8 = 8.0015 \)
5. \( (a^2 - 6a + 9) - (12 - 6a) = a^2 - 6a + 9 - 12 + 6a = a^2 - 3 \). При \( a=0.5 \): \( (0.5)^2 - 3 = 0.25 - 3 = -2.75 \)
6. \( -4p - p^2 + p^2 - 4 = -4p - 4 \).

Ответ: 1) 17 2) 0 3) -230.4 4) 8.0015 5) -2.75 6) -4p - 4