Ta có :  \(M=\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left[\left(x+1\right)\left(x+4\right)\right].\left[\left(x+2\right)\left(x+3\right)\right]+1=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

Đặt \(t=x^2+5x+5\) \(\Rightarrow M=\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\)

Vậy \(M=\left(x^2+5x+5\right)^2\)

(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)-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)

1) x² -7x + 10 = x² - 2x - 5x + 10 = x(x - 2) - 5(x - 2) = (x - 5)(x - 2)

2) x² + 3x + 2 = x² + 2x + x  + 2 = x(x + 2) + (x + 2) = (x + 1)(x + 2)

3) x² - 7x + 12 = x² - 3x - 4x + 12 = x(x - 3) - 4(x - 3) = (x - 3)(x - 4)

4) x² + 7x + 12 = x² + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4)

5) 16x - 5x² - 3 = 15x - 5x² + x - 3 = -5x(x - 3) + (x - 3) = (x - 3)(1 - 5x) 

6) 6x² + 7x - 3 = 6x² - 2x + 9x - 3 = 2x(3x - 1) + 3(3x - 1) = (2x + 3)(3x - 1)  

7) 3x² - 3x - 6 = 3x² - 6x + 3x - 6 = 3x(x - 2) + 3(x - 2) = (x - 2)(3x + 3) = 3(x - 2)(x + 1)

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

9) 6x² - 13x + 6 = 6x² - 9x -  4x + 6 = 3x(2x - 3) - 2(2x - 3) = (3x - 2)(2x - 3) 

10) 6x² + 15x  + 6 = 6x² + 12x + 3x + 6 = 6x(x + 2) + 3(x + 2) = (x + 2)(6x + 3) = 3(x + 2)(3x + 1)

11) 6x² - 20x + 6 = 6x² - 18x - 2x + 6 = 6x(x -3) - 2(x - 3) = (6x - 2)(x - 3) = 2(3x - 1)(x - 3)

12) 8x² + 5x - 3 = 8x² + 8x - 3x - 3 = 8x(x + 1) - 3(x + 1) = (x + 1)(8x - 3)

\(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-20\)

\(=\left(x^2+5x+4\right)\cdot\left(x^2+5x+6\right)-20\)

Đặt:   \(x^2+5x+5=a\)Khi đó ta có:

\(A=\left(a-1\right)\left(a+1\right)-20=a^2-21=\left(a-\sqrt{21}\right)\left(a+\sqrt{21}\right)\)

tự thay trở lại