Bài 2: Chứng minh: a) (x – 1)(x2 + x + 1) = x3 – 1

Bài 3:

a) (x + 1).(1 + x – x² + x³ – x⁴) – (x – 1).(1 + x + x² + x³ + x⁴)

= x + x² – x³ + x⁴ – x⁵ + 1 + x – x² + x³ – x⁴ – (x + x² + x³ + x⁴ + x⁵ – 1 – x – x² – x³ – x⁴)

= x + x² – x³ + x⁴ – x⁵ + 1 + x – x² + x³ – x⁴ – x - x² - x³ - x⁴ - x⁵ + 1 + x + x² + x³ + x⁴

= (x + x – x + x) + (x² – x² - x² + x²) + (– x³+ x³ - x³ + x³) + (x⁴– x⁴ - x⁴ + x⁴) + (– x⁵ - x⁵) + 1 + 1

= 2x – 2x⁵ + 2

b) (2b² – 2 – 5b + 6b³)(3 + 3b² – b)

= 6b² + 6b⁴ – 2b³ – 6 – 6b² + 2b – 15b – 15b³ + 5b² + 18b³ + 18b⁵ – 6b⁴

= (6b² – 6b² + 5b²) + (6b⁴ – 6b⁴) + (– 2b³ – 15b³ + 18b³) + (2b – 15b) + 18b⁵ – 6

= 5b² + b³ – 13b + 18b⁵ - 6

c) (4a – 4a² + 2a⁷)(6a² – 12 – 3a³)

= 24a³ – 28a – 12a⁴ – 24a⁴ + 28a² + 12a⁵ + 12a⁹ – 24a⁷ – 6a¹⁰

= -6a¹⁰ + 12a⁹ – 24a⁷ + 12a⁵ – 36a⁴ + 24a³ – 28a

d) (2ab + 2a² + b²)(2ab² + 4a³ – 4a²b)

= 4a²b³ + 8a⁴b – 8a³b² + 4a³b² + 8a⁵ – 8a⁴b + 2ab⁴ + 4a³b² – 4a²b³

= (4a²b³– 4a²b³) + (8a⁴b– 8a⁴b) + (4a³b²+ 4a³b²– 8a³b²) + 8a⁵ + 2ab⁴

= 0 + 0 + 0 + 8a⁵+ 2ab⁴

= 8a⁵ + 2ab⁴

Bài 4: Viết các biểu thức sau dưới dạng đa thức:

a) (2a – b)(b + 4a) + 2a(b – 3a)

= 2ab + 8a² – b² – 4ab + 2ab – 6a²

= 2a² – b²

b) (3a – 2b)(2a – 3b) – 6a(a – b)

= 6a² – 9ab – 4ab + 2b² – 6a² + 6ab

= 8a² – 7ab + 2b²

c) 5b(2x – b) – (8b – x)(2x – b)

= 10bx – 5b² – 16bx + 8b² + 2x² – bx

= 7bx + 3b² + 2x²

d) 2x(a + 15x) + (x – 6a)(5a + 2x)

= 2ax + 30x² + 5ax + 2x² – 30a² – 12ax

= 32x² – 30a² – 10ax

Hướng dẫn giải

Bài 2:

a) (x – 1)(x² + x + 1) = x³ – 1

Biến đổi vế trái ta có:

(x – 1)(x² + x + 1)

= x³ + x² + x – x² – x – 1

= x³ + (x² – x²) + (x – x) – 1

= x³ + 0x² + 0x – 1

= x³ – 1 = VP

b) (x³ + x²y + xy² + y³)(x – y) = x⁴ – y⁴

Biến đổi vế trái ta có:

(x³ + x²y + xy² + y³)(x – y)

= x⁴ – x³y + x³y – x²y² + x²y² – xy³ + y³x – y⁴

= (x⁴ – y⁴) + (– x³y + x³y) + (– x²y² + x²y²) + (– xy³ + y³x)

= x⁴ – y⁴ + 0. x³y + 0. x²y² + 0. xy³

= x⁴ – y⁴ = VP