x²-7x+12

=x²-4x-3x+12

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

=(x-3)(x-4)

        \(x^2-7x+12\)

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

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

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

\(12x^2+7x-12=12x^2-5x+12x-12\)

\(=x\left(12x-5\right)+12\left(x-1\right)\)

Đề sai rồi bạn ời

(x^2+3x+2)(x^2+7x+12)

=(x²+x+2x+2)(x²+3x+4x+12)

=[x.(x+1)+2.(x+1)][x.(x+3)+4.(x+3)]

=(x+1)(x+2)(x+3)(x+4)

(x^2+3x+2)(x^2+7x+12)+1

=(x²+x+2x+2)(x²+3x+4x+12)+1

=[x.(x+1)+2.(x+1)][x.(x+3)+4.(x+3)]+1

=(x+1)(x+2)(x+3)(x+4)+1

=[(x+1)(x+4)][(x+2)(x+3)]+1

=(x²+5x+4)(x²+5x+6)+1

=(x²+5x+4)[(x²+5x+4)+2]+1

=(x²+5x+4)²+2(x²+5x+4)+1

=(x²+5x+4+1)²

=(x²+5x+5)²

(x^2+3x+2)(x^2+7x+12)-24

=(x²+x+2x+2)(x²+3x+4x+12)-24

=[x.(x+1)+2.(x+1)][x.(x+3)+4.(x+3)]-24

=(x+1)(x+2)(x+3)(x+4)-24

=(x+1)(x+4)(x+2)(x+3)-24

=(x²+5x+4)(x²+5x+6)-24

Đặt t=x²+5x+4 ta được:

t.(t+2)-24

=t²+2t-24

=t²-4t+6t-24

=t.(t-4)+6.(t-4)

=(t-4)(t+6)

thay t=x²+5x+4 ta được:

(x²+5x+4-4)(x²+5x+4+6)

=(x²+5x)(x²+5x+10)

=x.(x+5)(x²+5x+10)

Vậy (x^2+3x+2)(x^2+7x+12)-24=x.(x+5)(x²+5x+10)

hay quá bạn ơi

Đúng nhưng hơi dài

Đặt x^2+5x+5=t thì ta được

(t-1)(t+1)-24=t^2-1-24=t^2-25=(t-1)(t+1)

=(x^2+5x)(x^2+5x+10)=x(x+5)(x^2+5x+10)

Sẽ gọn hơn

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)