quên rồi

lười quá ban

a) \(=\left(x^2-6\right)\left(x^2-1\right)=\left(x^2-6\right)\left(x-1\right)\left(x+1\right)\)

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

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

\(b,x^3-3x^2-4x+12\)

\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-4\right)\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

\(c,3x^3-7x^2+17x-5\)

\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)

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

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

\(\text{d) 2x}^4- 7x^3 - 2x^2 + 13x + 6\)
\(\text{= (2x^4 + 2x^3) - (9x^3 + 9x^2) + (7x^2 + 7x) + (6x + 6)}\)
\(\text{= 2x^3(x + 1) - 9x^2(x + 1) + 7x(x + 1) + 6(x + 1)}\)
\(\text{= (x + 1)(2x^3 - 9x^2 + 7x + 6)}\)
\(\text{= (x + 1)(2x + 1)(x - 3)(x - 2)}\)

phần b,c thay ''<=>'' là ''='' nhé ! Mình nhầm!

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)

a,\(3x^2-30x+75=3\left(x^2-10x+25\right)=3\left(x-5\right)^2\)

b, \(xy-x^2-x+y=x\left(y-x\right)+\left(y-x\right)=\left(y-x\right)\left(x+1\right)\)

c,\(x^2-7x-8=x^2-8x+x-8=x\left(x-8\right)+\left(x-8\right)=\left(x-8\right)\left(x+1\right)\)

a) 3x² - 30x + 75 = 3.(x² - 10x + 25) = 3.(x² - 2.5.x + 5²) = 3.(x-5)²

b) xy - x² - x + y = x.(y-x) + (y-x) = (y-x).(x+1)

c) x² - 7x - 8 = x² + x - 8x - 8 = x.(x+1) - 8.(x+1) = (x+1).(x-8)

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

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

d) \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

e) \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)

b: \(3x^2+2x-5\)

\(=3x^2-3x+5x-5\)

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

c: \(3-2x-x^2\)

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

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

d: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

e: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)

a) Ta có : x² + 7x + 12 

= x² + 3x + 4x + 12

= (x² + 3x) + (4x + 12)

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

= (x + 4)(x + 3)

Bạn ơi mk nhầm đề rồi số 30 thay bằng số 60 còn 36 thay bằng 72 và 39 thay bằng 75 nha 

Ta có \(x^2+7x+12\)

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

=\(\left(x^2+3x\right)+\left(4x+12\right)\)

=\(x\left(x+3\right)+4\left(x+3\right)\)

=\(\left(x+4\right)\left(x+3\right)\)