a, x^2 + 5x +4

= x^2 + 1x + 4x + 4

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

= x ( x + 1 ) + 4 ( x + 1 )

= (x + 1) (x + 4)

b, x^2 - 6x + 5

= x^2 - 1x - 5x + 5

= (x^2 - 1x) - (5x - 5)

= x (x - 1) - 5 (x - 1)

= (x - 1) (x - 5)

c, x^2 + 7x + 12

= x^2 + 3x + 4x + 12 

= (x^2 + 3x) + (4x + 12)

= x (x + 3) + 4 (x + 3)

= (x + 3) (x + 4)

d, 2x^2 - 5x + 3

= 2^x2 - 2x - 3x + 3

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

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

e, 7x  - 3x^2 - 4

= 3x + 4x - 3x^2 - 4

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

= 3x (1 - x) + 4 (x - 1)

= 3x (1-x) - 4 (1 - x)

= (1 - x) (3x - 4)

f, x^2 - 10x + 16

= x^2 - 2x - 8x + 16

= (x^2 - 2x) - (8x - 16)

= x (x - 2) - 8 (x - 2)

= (x - 2) (x - 8)

a, (x+1)(x+4)

b,(x-5)(x-1)

c,(x+3)(x+4)

d,(2x-3)(x-1)

e,(-3x+4)(x-1)

f, (x-8)(x-2)

1) \(x^2-y^2-2x-2y\)

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

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

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

2) \(3x^2-3y^2-2\left(x-y\right)^2\)

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

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

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

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

\(=\left(x-y\right)\left(x+5y\right)\)

1) x² - y² - 2x - 2y

= (x² - y²) - (2x + 2y)

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

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

2) 3x² - 3y² - 2(x - y)²

= (3x² - 3y²) - 2(x - y)²

= 3(x² - y²) - 2(x - y)²

= 3(x - y)(x + y) - 2(x - y)²

= (x - y)[3(x + y) - 2(x - y)]

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

= (x - y)(x + 5y)

`x^2-y^2 -2x-2y`

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

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

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

__

`3x^2 -3y^2 -2(x-y)^2`

`= 3(x^2 -y^2) - 2(x-y)^2`

`=3(x-y)(x+y) -2(x-y)^2`

`= (x-y) (3x+3y -2x+2y)`

`=(x-y)( x+5y)`

a, Cách 1 : \(x^2+5x+6=x^2+2x+3x+6=\left(x+2\right)\left(x+3\right)\)

Cách 2 : \(x^2+5x+6=x^2+2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}+6\)

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

b, Cách 1 : \(x^2-x-6=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\)

Cách 2 : \(x^2-x-6=x^2-x+\frac{1}{4}-\frac{1}{4}-6=\left(x-\frac{1}{2}\right)^2-\frac{25}{4}=\left(x-3\right)\left(x+2\right)\)

c, Cách 1 : \(x^2+6x+8=x^2+4x+2x+8=\left(x+2\right)\left(x+4\right)\)

Cách 2 : \(x^2+6x+8=x^2+6x+9-1=\left(x+3\right)^2-1=\left(x+2\right)\left(x+4\right)\)

d, Cách 1 : \(x^2-2x-8=x^2+2x-4x-8=\left(x-4\right)\left(x+2\right)\)

Cách 2 : \(x^2-2x-8=x^2-2x+1-9=\left(x-1\right)^2-9=\left(x-4\right)\left(x+2\right)\)

1: \(\dfrac{A}{B}=\dfrac{2x^4+4x^3-x^3-2x^2-2x^2-4x+x+2}{x+2}\)

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

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

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

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

(x+1)(x-4)(x+2)(x-8)+4x^2

=[(x+1)(x-8)][(x-4)(x+2)]+4x²

=(x²-7x-8)(x²-2x-8)+4x²

Đặt t=x²-2x-8 ta được:

(t-5x).t+4x²

=t²-5xt+4x²

=t²-xt-4xt+4x²

=t.(t-x)-4x.(t-x)

=(t-x)(t-4x)

thay t=x²-2x-8 ta được:

(x²-3x-8)(x²-6x-8)

Vậy (x+1)(x-4)(x+2)(x-8)+4x^2=(x²-3x-8)(x²-6x-8)

Câu 1: C

Câu 2: A

Câu 3: B

Câu 4:B

Câu 5: A

a: \(x^3-2x+4\)

\(=x^3+2x^2-2x^2-4x+2x+4\)

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

b: \(x^3-4x^2+12x-27\)

\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\left(x-3\right)\)

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

c: \(x^3+2x^2+2x+1\)

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

\(=\left(x+1\right)\left(x^2+x+1\right)\)

Đặt x^2 + 2x = y thay vào ta có:

 y(y+4) + 3 = y^2 + 4y +3 = y^2 + y + 3y + 3 = y(y+1) + 3(y + 1) = ( y + 3)( y+ 1)

Thay y = x^2 + 2x ta có

 ( x^2 + 2x + 3)(x^2 + 2x+ 1) = ( x^2 + 2x + 3) (x+ 1)^2

Đúng cho mình nha

\(\left(x^2+2x\right)\left(x^2+2x+4\right)+3\)

Đặt \(x^2+2x+2=t\)

\(\Rightarrow\left(t-2\right)\left(t+2\right)+3=t^2-4+3=t^2-1=\left(t-1\right)\left(t+1\right)\)

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

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

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

mk chỉ phân tích thôi bạn tự chia nha!
a, \(16x^4-81=(4x^2)^2-9^2=(4x^2-9)(4x^2+9)\)

                    \(=[(2x)^2-3^2](4x^2+9)\)

                    \(=(2x+3)(2x-3)(4x^2+9)\)

b, \(x^3-3x^2+3x-1=(x-1)^3\)

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

c, \(18x^5+9x^4+3x^3+6x^2+3x+1=(18x^5+9x^4+3x^3)+(6x^2+3x+1)\)

\(=(6x^2+3x+1)(3x^3+1)\)

câu c bạn đánh sai 1 dấu phép toán kìa!!!!