b) (2x + y)2 = (2x)2 + 2 . 2x . y + y2 = 4x2 + 4xy + y2;

c) (3 – x)2 = 32 – 2 . 3 . x + x2;

d) (x – 4y)2 = x2 – 2 . x . 4y + (4y)2 = x2 – 8xy + 16y2.

a) $y^2+y+\frac{1}{4}=y^2+2.\frac{1}{2}y+{\left(\frac{1}{2}\right)}^2={\left(y+\frac{1}{2}\right)}^2$;

b) y2 + 49 – 14y = y2 – 2 . 7 . y + 72 = (y – 7)2.

Ta có 492 = (50 – 1)2 = 502 – 2 . 50 . 1 + 12

= 2500 – 100 + 1 = 2400 + 1 = 2401.

Ta có: (a – b)(a + b) = a . a + a .  b – b . a + b . b = a2 – b2.

a) 9x2 – 16 = (3x)2 – 42 = (3x + 4)(3x – 4);

b) 25 – 16y2 = 52 – (4y)2 = (5 + 4y)(5 – 4y).

b) (2x + 5)(2x – 5) = (2x)2 – 52 = 4x2 – 25;

c) (4y – 1)(4y + 1) = (4y)2 – 1 = 16y2 – 1.

Ta có: 48 . 52 = (50 – 2)(50 + 2) = 502 – 22 = 2500 – 4 = 2496.

a) (a + b)(a + b)2 = (a + b)(a2 + 2ab + b2)

= a . a2 + a . 2ab + a . b2 + b . a2 + b . 2ab + b . b2

= a3 + 2a2b + ab2 + a2b + 2ab2 + b3

= a3 + 3a2b + 3ab2 + b3.

b) (a – b)(a – b)2 = (a – b)(a2 – 2ab + b2)

= a . a2 – a . 2ab + a . b2 – b . a2 + b . 2ab – b .  b2

= a3 – 2a2b + ab2 – a2b + 2ab2 – b3

= a3 – 3a2b + 3ab2 – b3.

a) (3 + x)3 = 33 + 3 . 32 . x + 3 . 3 . x2 + x3 = 27 + 27x + 9x2 + x3;

b) (a + 2b)3 = a3 + 3 . a2 . 2b + 3 . a . (2b)2 + (2b)3

= a3 + 6a2b + 12ab2 + 8b3;

c) (2x – y)3 = 2x3 – 3 . (2x)2 . y + 3 . 2x . y2 – y3

= 2x3 – 12x2y + 6xy2 – y3.

Ta có: 8x3 – 36x2y + 54xy2 – 27y3

= (2x)3 – 3 . (2x)2 . 3y + 3 . 2x . (3y)2 – (3y)3

= (2x – 3y)3.

101³ – 3 . 101² + 3 . 101 – 1

= 101³ – 3 . 101² . 1 + 3 . 101 . 1² – 1³

= (101 – 1)³

= 100³

= 1 000 000.

a) (a + b)(a2 – ab + b2) = a . a2 – a . ab + a . b2 + b . a2 – b . ab + b . b2

= a3 – a2b + ab2 + a2b – ab2 + b3 = a3 + b3.

b) (a – b)(a2 + ab + b2) = a . a2 + a . ab + a . b2 – b . a2 – b . ab – b . b2

= a3 + a2b + a2b – a2b – a2b – b3 = a3 – b3.

a) 27x3 + 1 = (3x)3 + 13 = (3x + 1)[(3x)2 – 3x . 1 + 12]

= (3x + 1)(9x2 – 3x + 1);

b) 64 – 8y3 = 43 – (2y)3 = (4 + 2y)(4 – 2y).

a) 4x2 + 28x + 49 = (2x)2 + 2 . 2x . 7 + 72 = (2x + 7)2;

b) 4a2 + 20ab + 25b2 = (2a)2 + 2 . 2a . 5b + (5b)2 = (2a + 5b)2;

c) 16y2 – 8y + 1 = (4y)2 – 2 . 4y . 1 + 12 = (4y – 1)2;

d) 9x2 – 6xy + y2 = (3x)2 – 2 . 3x . y + y2 = (3x – y)2.

a) a³ +12a² + 48a + 64

= a³ + 3.a².4 + 3.a.4² + 4³

= (a + 4)³.

b) 27x3 + 54x2y + 36xy2 + 8y3

= (3x)3 + 3 . (3x)2 . 2y + 3 . 3x . (2y)2 + (2y)3

= (3x + 2y)3;

c) x3 – 9x2 + 27x – 27 = x3 – 3 . x2 . 3 + 3 . x . 32 – 33 = (x – 3)3;

d) 8a3 – 12a2b + 6ab2 – b3 = (2a)3 – 3 . (2a)2b + 3 . 2ab2 – b3 = (2a – b)3.

a) 25x2 – 16 = (5x)2 – 42 = (5x + 4)(5x – 4);

b) 16a2 – 9b2 = (4a)2 – (3b)2 = (4a + 3b)(4a – 3b);

c) 8x3 + 1 = (2x)3 + 1 = (2x + 1)[(2x)2 + 2x . 1 + 12] = (2x + 1)(4x2 + 2x + 1);

d) 125x3 + 27y3 = (5x)3 + (3y)3 = (5x + 3y)[(5x)2 + 5x . 3y + (3y)2]

= (5x + 3y)(25x2 + 15xy + 9y2);

e) 8x3 – 125 = (2x)3 – 53 = (2x + 5)[(2x)2 + 2x . 5 + 52]

= (2x + 5)(4x2 + 10x + 25);

g) 27x3 – y3 = (3x)3 – y3 = (3x + y)(3x – y).

a) Ta có A = x2 + 6x + 10 = x2 + 6x + 9 + 1 = (x + 3)2 + 1.

Thay x = −103 vào biểu thức A, ta được:

A = (−103 + 3)2 + 1 = (−100)2 + 1 = 10 000 + 1 = 10 001.

Vậy A = 10 001 tại x = −103.

b) Ta có B = x3 + 6x2 + 12x + 12 = x3 + 3 . x2 . 2 + 3 . x . 22 + 23 + 4

= (x + 2)3 + 4.

Thay x = 8 vào biểu thức B, ta được:

B = (8 + 2)3 + 4 = 103 + 4 = 1004.

Vậy B = 1004 tại x = 8.

a) Ta có C = (3x – 1)2 + (3x + 1)2 – 2(3x – 1)(3x + 1)

= [(3x – 1) – (3x + 1)]2 = (3x – 1 – 3x – 1)2

= (– 1 – 1)2 = (–2)2 = 4.

Vậy biểu thức C không phụ thuộc vào biến x.

b) D = (x + 2)3 – (x – 2)3 – 12(x2 + 1)

= [(x + 2) – (x – 2)][(x + 2)2 + (x + 2)(x – 2) + (x – 2)2] – 12(x2 + 1)

= (x + 2 – x + 2)[(x + 2)2 + x2 – 22 + (x – 2)2] – 12x2 – 12

= 4(x2 + 4x + 4 + x2 – 4 + x2 – 4x + 4) – 12x2 – 12

= 4(3x2 + 4) – 12x2 – 12

= 12x2 + 16 – 12x2 – 12 = 4.

Vậy biểu thức D không phụ thuộc vào biến x.

c) E = (x + 3)(x2 – 3x + 9) – (x – 2)(x2 + 2x + 4)

= (x3 + 33) – (x3 – 23) = x3 + 27 – x3 + 8 = 35.

Vậy biểu thức E không phụ thuộc vào biến x.

d) G = (2x – 1)(4x2 + 2x + 1) – 8(x + 2)(x2 – 2x + 4)

= [(2x)3 – 13] – 8(x3 + 23) = (8x3 – 1) – 8(x3 + 8)

= 8x3 – 1 – 8x3 – 64 = – 65.

Vậy biểu thức D không phụ thuộc vào biến x.