đây a 

  day ạ

Câu 44: 

\(12x-9-4x^2=-4x^2+12x-9=-\left(4x^2-12x+9\right)\)

\(=\left[\left(2x\right)^2-2\cdot3\cdot2x+3^2\right]=-\left(2x-3\right)^2\)

⇒ Chọn C

Câu 45: 

\(x^3-6x^2y+12xy^2-8y^3=x^3-3\cdot2y\cdot x^2+3\cdot\left(2y\right)^2-\left(2y\right)^3\)

\(=\left(x-2y\right)^3\)

⇒ Chọn D

5

\(=4^2-x^2=\left(4-x\right)\left(4+x\right)\)

6

\(=4^2-\left(3x+1\right)^2=\left(4-3x-1\right)\left(4+3x+1\right)=\left(3-3x\right)\left(5+3x\right)\\ =3\left(1-x\right)\left(5+3x\right)\)

7

\(=\left(2x+5\right)^2-\left(3x\right)^2=\left(2x+5+3x\right)\left(2x+5-3x\right)\\ =\left(5x+5\right)\left(5-x\right)\\ =5\left(x+1\right)\left(5-x\right)\)

8

\(=\left(2x-1-3x+1\right)\left(2x-1+3x-1\right)=\left(-x\right)\left(5x-2\right)\)

9

\(=\left(2x\right)^2-2.2x.y+y^2=\left(2x-y\right)^2\)

10

\(=\left(x+1\right)^2-\left(3y\right)^2=\left(x+1-3y\right)\left(x+1+3y\right)\)

11

\(=\left(x^2y^2\right)^2+2.2x^2y^2+2^2=\left(x^2y^2+2\right)\)

12

\(=\left(y-2\right)^2-x^2=\left(y-2-x\right)\left(y-2+x\right)\)

13

\(=1-\left(3\sqrt{3}x\right)^3=\left(1-3\sqrt{3}x\right)\left[1^2+3\sqrt{3}.x+\left(3\sqrt{3}.x\right)^2\right]=\left(1-3\sqrt{3}x\right)\left(1+3\sqrt{3}x+27x^2\right)\)

  đây

  làm 1 bài cx đc hết càng tốt ạ

Bài 5:

a) \(M=\left(2x-1\right)^2+2\left(2x-1\right)\left(3x+1\right)+\left(3x+1\right)^2\)

\(M=\left[\left(2x-1\right)+\left(3x+1\right)\right]^2\)

\(M=\left(5x\right)^2\)

Thay \(x=-\dfrac{1}{5}\) vào biểu thức M ta có:

\(\left(5\cdot-\dfrac{1}{5}\right)^2=\left(-1\right)^2=1\)

Vậy: ...

b) \(N=\left(3x-1\right)^2-2\left(9x^2-1\right)+\left(3x+1\right)^2\)

\(N=\left(3x-1\right)^2-2\left(3x+1\right)\left(3x-1\right)+\left(3x+1\right)^2\)

\(N=\left[\left(3x-1\right)-\left(3x+1\right)\right]^2\)

\(N=\left(-2\right)^2=4\)

Vậy: ...

4

a

\(=2xy\left(x^2+y^2+2xy-1\right)=2xy\left[\left(x+y\right)^2-1^2\right]\\ =2xy\left(x+y-1\right)\left(x+y+1\right)\)

b

\(=\left(x-y\right)^2+4\left(x-y\right)=\left(x-y\right)\left(x-y+4\right)\)

c

\(=\left(x+y\right)^3-\left(x+y\right)\\ =\left(x+y\right)\left[\left(x+y\right)^2-1\right]=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)

d

\(=\left(x-y\right)^2-\left(2z\right)^2=\left(x-y-2z\right)\left(x-y+2z\right)\)

e

\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)

f

\(=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)

  đây ạ

14: =x^3-3x^2+3x-1-1

=(x-1)^3-1

=(x-1-1)(x^2-2x+1+x-1+1)

=(x-2)(x^2-x+1)

15: =x^3+x^2-4x^2-4x+7x+7

=(x+1)(x^2-4x+7)

16: =2x^3-x^2-2x^2+x+2x-1

=(2x-1)(x^2-x+1)

Đây ạ  

1: =x^2-2*x*2y+4y^2

=(x-2y)^2

2: 9x^2-y^2=(3x-y)(3x+y)

5: x^3-8=(x-2)(x^2+2x+4)

4: =x^2-(y-2)^2

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

7: 8x^3+12x^2+6x+1=(2x+1)^3

14: =x^3-3x^2+3x-1-1

=(x-1)^3-1

=(x-1-1)(x^2-2x+1+x-1+1)

=(x-2)(x^2-x+1)

15: =x^3+x^2-4x^2-4x+7x+7

=(x+1)(x^2-4x+7)

16: =2x^3-x^2-2x^2+x+2x-1

=(2x-1)(x^2-x+1)

a) (x + 2)³ + 1

= (x + 2)³ + 1³

= [(x + 2) + 1][(x + 2)² - (x + 2).1 + 1²]

= (x + 3)(x² + 4x + 4 - x - 2 + 1)

= (x + 3)(x² + 3x + 3)

b) x³ + 6x² + 12x + 9

= x³ + 3x² + 3x² + 9x + 3x + 9

= (x³ + 3x²) + (3x² + 9x) + (3x + 9)

= x²(x + 3) + 3x(x + 3) + 3(x + 3)

= (x + 3)(x² + 3x + 3)

c) x³ + 6x² + 12x + 7

= x³ + x² + 5x² + 5x + 7x + 7

= (x³ + x²) + (5x² + 5x) + (7x + 7)

= x²(x + 1) + 5x(x + 1) + 7(x + 1)

= (x + 1)(x² + 5x + 7)

d) 2x³ + 6x² + 12x + 8

= 2(x³ + 3x² + 6x + 4)

= 2(x³ + x² + 2x² + 2x + 4x + 4)

= 2[(x³ + x²) + (2x² + 2x) + (4x + 4)]

= 2[x²(x + 1) + 2x(x + 1) + 4(x + 1)]

= 2(x + 1)(x² + 2x + 4)

a: =(x+2+1)(x^2+4x+4-x-2+1)

=(x+3)(x^2+3x+3)

b: =x^3+6x^2+12x+8+1

=(x+2)^3+1

=(x+3)(x^2+3x+3)

c: =x^3+x^2+5x^2+5x+7x+7

=(x+1)(x^2+5x+7)

  làm VD2 ạ

VD2:

a: \(3-x^2=\left(\sqrt{3}-x\right)\left(\sqrt{3}+x\right)\)

b: 3-(x+1)^2

\(=\left(\sqrt{3}-x-1\right)\left(\sqrt{3}+x+1\right)\)

c: (x+5)^2-4x^2

=(x+5+2x)(x+5-2x)

=(-x+5)(3x+5)

d: (x+1)^2-(2x-1)^2

=(x+1+2x-1)(x+1-2x+1)

=(2-x)*3x

Bài 6: 

a) \(P=27-27a+9x^2-x^3\)

\(P=3^3-3\cdot3^2\cdot x+3\cdot3\cdot x^2-x^3\)

\(P=\left(3-x\right)^3\)

Thay \(x=-17\) vào P ta được:

\(\left(3--17\right)^3=20^3=8000\)

Vậy: ...

b) \(Q=x^3+3x^2+3x\)

\(Q=x^3+3x^2+3x+1-1\)

\(Q=\left(x+3\right)^3-1\)

Thay \(x=99\) vào Q ta có:

\(\left(99+1\right)^3-1=100^3-1=1000000-1=999999\)

Vậy: ...

  bài 6 ạ

6:

a: P=(3-x)^3=(3+17)^3=20^3=8000

b: Q=x^3+3x^2+3x+1-1

=(x+1)^3-1

=(99+1)^3-1

=100^3-1

=999999