giải bài này giúp mih với ạ

Bài 1. 1) $x^2 + 8x + 16 = (x + 4)^2$ 2) $9x^2 - 6x + 1 = (3x - 1)^2$ 3) $10x - 25 - x^2 = -(x^2 - 10x + 25) = -(x - 5)^2$ 4) $x^2 + 5x + \frac{25}{4} = (x + \frac{5}{2})^2$ 5) $16 - x^2 = (4 - x)(4 + x)$ 6) $16 - (3x + 1)^2 = (4 - (3x + 1))(4 + (3x + 1)) = (3 - 3x)(5 + 3x)$ 7) $(2x + 5)^2 - 9x^2 = ((2x + 5) - 3x)((2x + 5) + 3x) = (-x + 5)(5x + 5)$ 8) $(2x - 1)^2 - (3x - 1)^2 = ((2x - 1) - (3x - 1))((2x - 1) + (3x - 1)) = (-x)(5x - 2)$ 9) $x^2 + 4xy + y^2 = (x + 2y)^2$ 10) $(x + 1)^2 - 9y^2 = ((x + 1) - 3y)((x + 1) + 3y) = (x - 3y + 1)(x + 3y + 1)$ 11) $x^4y^4 + 4x^2y^2 + 4 = (x^2y^2 + 2)^2$ 12) $y^2 - 4y + 4 - x^2 = (y - 2)^2 - x^2 = ((y - 2) - x)((y - 2) + x) = (y - x - 2)(y + x - 2)$ 13) $1 - 27x^3 = (1 - 3x)(1 + 3x + 9x^2)$ 14) $(x - 3)^3 + 27 = (x - 3 + 3)((x - 3)^2 - 3(x - 3) + 9) = x(x^2 - 6x + 9 - 3x + 9 + 9) = x(x^2 - 9x + 27)$ 15) $27x^3 + 27x^2 + 9x + 1 = (3x + 1)^3$ 16) $\frac{x^6}{27} - \frac{x^4y}{3} + x^2y - y = y(\frac{x^6}{27} - \frac{x^4}{3} + x^2 - 1) = y((\frac{x^2}{3})^3 - (\frac{x^2}{3})^2 + \frac{x^2}{3} - 1)$ 17) $(2x - 1)^3 + 8 = (2x - 1 + 2)((2x - 1)^2 - 2(2x - 1) + 4) = (2x + 1)(4x^2 - 4x + 1 - 4x + 2 + 4) = (2x + 1)(4x^2 - 8x + 7)$ 18) $8x^3 - 12x^2 + 6x - 1 = (2x - 1)^3$ 19) $8x^3 - 12x^2 + 6x - 2 = 2(4x^3 - 6x^2 + 3x - 1) = 2((2x - 1)^3)$ 20) $9x^3 - 12x^2 + 6x - 1 = (3x - 1)^3$