Đáp án + Giải thích các bước giải:

Bài $2$:

$\textbf{1$\bigg)$} 2(x - y)(x + y) + (x - y)^2 + (x + y)^2$

$= (x - y)^2 + 2(x - y)(x + y) + (x + y)^2$

$= [(x - y) + (x + y)]^2$

$= (x - y + x + y)^2$

$= (2x)^2$

$= 4x^2$

$\textbf{2$\bigg)$} (x - y + z)^2 + (z - y)^2 + 2(x - y + z)(y - z)$

$= (x - y + z)^2 + 2(x - y + z)(y - z) + [-(y - z)]^2$

$= (x - y + z)^2 + 2(x - y + z)(y - z) + (y - z)^2$

$= [(x - y + z) + (y - z)]^2$

$= (x - y + z + y - z)^2$

$= x^2$

$\textbf{3$\bigg)$} (a + 1)(a + 2)(a^2 + 4)(a - 1)(a^2 + 1)(a - 2)$

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

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

$= [(a^2)^2 - 1][(a^2)^2 - 16]$

$= (a^4 - 1)(a^4 - 16)$

$= a^8 - a^4 - 16a^4 + 16$

$= a^8 - 17a^4 + 16$

$\textbf{4$\bigg)$} (a + 2b - 3c - d)(a + 2b + 3c + d)$

$= [a + 2b - (3c + d)](a + 2b + 3c + d)$

$= (a + 2b)^2 - (3c + d)^2$

$= a^2 + 4ab + 4b^2 - (9c^2 + 6cd + d^2)$

$= a^2 + 4ab + 4b^2 - 9c^2 - 6cd - d^2$

$\textbf{5$\bigg)$} (1 - x - 2x^3 + 3x^2)(1 - x + 2x^3 - 3x^2)$

$= [1 - x - (2x^3 - 3x^2)](1 - x + 2x^3 - 3x^2)$

$= (1 - x)^2 - (2x^3- 3x^2)^2$

$= 1 - 2x + x^2 - [x^2(2x - 3)]^2$

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

$= 1 - 2x + x^2 - x^4(4x^2 - 12x + 9)$

$= 1 - 2x + x^2 - 4x^6 + 12x^5 - 9x^4$

$= -4x^6 + 12x^5 - 9x^4 + x^2 - 2x + 1$

$\textbf{6$\bigg)$} (a^6 - 3a^3 + 9)(a^3 + 3)$

$= [(a^3)^2 - 3a^3 + 9](a^3 + 3)$

$= (a^3)^3 + 3^3$

$= a^9 + 27$

$\textbf{7$\bigg)$} (3x + 1)^2 - 2(3x + 1)(3x + 5) + (5x + 5)^2$

$= (3x + 1)^2 - 2(3x + 1)(3x + 5) + (3x + 5)^2 + (5x + 5)^2 - (3x + 5)^2$

$= [(3x + 1) - (3x+ 5)]^2 + [(5x + 5) + (3x + 5)][(5x + 5) - (3x + 5)]$

$= (3x + 1 - 3x - 5)^2 + (5x + 5 + 3x + 5)(5x + 5 - 3x - 5)$

$= 16 + 2x(8x + 10)$

$= 16x^2 + 20x + 16$