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

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

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

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

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

3: \(=2\left(x+2\right)\left(25x-15-x\right)\)

\(=2\left(x+2\right)\left(24x-15\right)\)

=6(x+2)(8x-5)

(12x^2 - 12xy + 3y^2) - 10.(2x - y) + 8 

= 3(4x^2 - 4xy + y^2) - 10(2x - y) + 8 

= 3(2x - y)^2 - 10(2x - y) + 8 

= 3(2x - y)^2 - 10(2x - y) + 8 

= 3(2x - y)^2 - 6(2x - y) - 4(2x - y) + 8 

= 3(2x - y)(2x - y - 2) - 4(2x - y -2) 

= (2x - y -2)[3(2x - y) - 4] 

= (2x - y -2)(6x - 3y -4)

Ai k mk mk k lại

(12x^2 - 12xy + 3y^2) - 10.(2x - y) + 8 
= 3(4x^2 - 4xy + y^2) - 10(2x - y) + 8 
= 3(2x - y)^2 - 10(2x - y) + 8 
= 3(2x - y)^2 - 10(2x - y) + 8 
= 3(2x - y)^2 - 6(2x - y) - 4(2x - y) + 8 
= 3(2x - y)(2x - y - 2) - 4(2x - y -2) 
= (2x - y -2)[3(2x - y) - 4] 
= (2x - y -2)(6x - 3y -4)

a)  \(A=\left(x+2\right)\left(x+3\right)\left(x+5\right)\left(x+6\right)-10\)

\(=\left(x^2+8x+12\right)\left(x^2+8x+15\right)-10\)

Đặt   \(x^2+8x+12=t\)

Khi đó ta có: 

\(A=t\left(t+3\right)-10\)

   \(=t^2+3t-10\)

   \(=\left(t-2\right)\left(t+5\right)\)

Thay trở lại ta có:

\(A=\left(x^2+8x+10\right)\left(x^2+8x+17\right)\)

b)  \(B=x\left(2x+1\right)\left(2x+3\right)\left(4x+8\right)-18\)

\(=\left(4x^2+8x\right)\left(4x^2+8x+3\right)-18\)

Đặt  \(4x^2+8x=t\)

Khi đó ta có:

\(B=t\left(t+3\right)-18=t^2+3t-18=\left(t-3\right)\left(t+6\right)\)

Thay trở lại ta có:

\(B=\left(4x^2+8x-3\right)\left(4x^2+8x+6\right)=2\left(4x^2+8x-3\right)\left(2x^2+4x+3\right)\)

Mọi người đã hướng dẫn bạn cách làm rồi mà.

a, Đặt A=...=(x+2)(x+6)(x+3)(x+5)-10=(x²+8x+12)(x²+8x+15)-10

Đặt x²+8x+12=y

=>A=y(y+3)-10=y²+3y-10=y²-2y+5y-10=y(y-2)+5(y-2)=(y-2)(y+5)=(x²+8x+12-2)(x²+8x+12+5)=(x²+8x+10)(x²+8x+17)

b, Đặt B=...=x(4x+8)(2x+1)(2x+3)-18=(4x²+8x)(4x²+8x+3)-18

Đặt 4x²+8x=t

=>B=t(t+3)-18=t²+3t-18=t²-3t+6t-18=t(t-3)+6(t-3)=(t-3)(t+6)=(4x²+8x-3)(4x²+8x+6)

x^3+x^2-2x-8

= (x-2)(x^2+3x+4)

nah bạn chúc bạn học tốt nha 

     x³ + x² - 2x - 8

=  ( x³ - 8 ) + ( x² - 2x )

=  ( x - 2 ) . ( x² + 2x + 4 ) + x ( x - 2 )

= ( x - 2 ) .( x² + 2x + 4 + x )

= ( x-2 ) . ( x² + 3x + 4 ) 

x^3+x^2-2x-8

= (  x^3 - 8) + ( x^2 - 2x )

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

=(  x - 2)( x^2 + 2x + 4 + x)  

=(  x - 2)( x^2 + 3x + 4 )  

\(x^3-8+2x\left(x-2\right)\\ =\left(x-2\right)\left(x^2+2x+4\right)+2x\left(x-2\right)\\ =\left(x-2\right)\left(x^2+2x+4+2x\right)=\left(x-2\right)\left(x^2+4x+4\right)\\ =\left(x-2\right)\left(x+2\right)^2\)

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

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

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