a) \[ (2x-1)^2 = (2x)^2 - 2 \cdot 2x \cdot 1 + 1^2 = 4x^2 - 4x + 1 \]
б) \[ (3a+c)^2 = (3a)^2 + 2 \cdot 3a \cdot c + c^2 = 9a^2 + 6ac + c^2 \]
в) \[ (y-5)(y+5) = y^2 - 5^2 = y^2 - 25 \]
г) \[ (4b+5c)(4b-5c) = (4b)^2 - (5c)^2 = 16b^2 - 25c^2 \]
\[ (x+y)(x-y) - (x^2+3y^2) = x^2 - y^2 - x^2 - 3y^2 = -4y^2 \]
a) \[ 16y^2 - 0.25 = (4y)^2 - (0.5)^2 = (4y - 0.5)(4y + 0.5) \]
б) \[ a^2 + 10ab + 25b^2 = a^2 + 2 \cdot a \cdot 5b + (5b)^2 = (a + 5b)^2 \]
\[ (5-x)^2 - x(2.5+x) = 0 \] \[ 25 - 10x + x^2 - 2.5x - x^2 = 0 \] \[ 25 - 12.5x = 0 \] \[ 12.5x = 25 \] \[ x = \frac{25}{12.5} = 2 \]
a) \[ (2a-b^2)(2a+b^2) = (2a)^2 - (b^2)^2 = 4a^2 - b^4 \]
б) \[ (x-6x^2)^2 = x^2 - 2 \cdot x \cdot 6x^2 + (6x^2)^2 = x^2 - 12x^3 + 36x^4 \]
в) \[ (y+b)^2(y-b)^2 = ((y+b)(y-b))^2 = (y^2 - b^2)^2 = y^4 - 2y^2b^2 + b^4 \]
a) \[ \frac{1}{81}a^2 - 0.09c^4 = (\frac{1}{9}a)^2 - (0.3c^2)^2 = (\frac{1}{9}a - 0.3c^2)(\frac{1}{9}a + 0.3c^2) \]
б) \[ (b+8)^2 - 4b^2 = (b+8)^2 - (2b)^2 = (b+8 - 2b)(b+8 + 2b) = (8-b)(3b+8) \]
в) \[ a^3 - b^3 = (a-b)(a^2 + ab + b^2) \]
Ответ: См. решение выше