\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

(x² - x)² - 2 * (x² - x) - 15 

đặt x² - x = a

có: a² - 2a - 15 = (a² - 2a + 1) - 16 = (a - 1)² - 16 = (a - 5) (a + 3) 

thay vào đc:  (x² - x - 5) (x² - x +3)

đặt x^2 - x = t
phương trình trở thành 
t^2-2t-15
= t^2 - 2t + 1 -16
= (t^2 - 1) - 16
=> (x^2 - x - 1) -16

bài này có bạn hỏi ở dưới rồi đấy bạn!

(x^2+x)^2-2(x^2+x)-15

=(x²+x)²-2(x²+x)+1-16

=(x²+x-1)²-16

=(x²+x-1+4)(x²+x-1-4)

=(x²+x+3)(x²+x-5)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

( x² + 8x + 7 ) ( x² + 8x + 15 ) + 15

Đặt x² + 8x + 7 = y ta có:

   y ( y + 8 ) + 15 

= y² + 8y + 15 

= ( y + 3 ) ( y + 5 )

= ( x² + 8x + 10 ) ( x² + 8x + 12 ) 

= ( x² + 8x + 10 ) ( x + 2 ) ( x + 6 )

Đặt x² + 8x + 7 = y ta có:

   y ( y + 8 ) + 15 

= y² + 8y + 15 

= ( y + 3 ) ( y + 5 )

= ( x² + 8x + 10 ) ( x² + 8x + 12 ) 

= ( x² + 8x + 10 ) ( x + 2 ) ( x + 6 )

đặt t = x² + 8x + 7

có : t ( t + 8 ) + 15 

= t² + 8t + 15

= t² + 5t + 3t + 15

= ( t² + 5t ) + ( 3t + 15 )

= t ( t + 5 ) + 3 ( t + 5 )

= ( t + 3 ) ( t + 5 )

mà t = x² + 8x + 7

⇒ ( x² + 8x + 10 ) ( x² + 8x + 12 )

=(x^2+8x)^2+23(x^2+8x)+135

Cái này ko phân tích được nha bạn

\(\left(x^2+8x+8\right)\left(x^2+8x+15\right)+15\\ \Leftrightarrow\left(x^4+8x^3+15x^2+8x^3+64x^2+120x+8x^2+64x+120\right)+15\\ \Leftrightarrow x^4+16x^3+87x^2+184x+135\)

Gọi `A=(x^2+8x+8)(x^2+8x+15)+15`

Đặt `t=x^2+8x+11,5`

`=>A=(t-3,5)(t+3,5)+15=t^2-3,5^2+15=t^2-2,75=(t-sqrt(2,75))(t+sqrt(2,75))=(x^2+8x+11,5-(sqrt11)/2)(x^2+8x+11,5+(sqrt11)/2)=(x^2+8x+(23-\sqrt11)/2)(x^2+8x+(23+\sqrt11)/2)`

mở sách giải ra mà cop

\(x^4+x^3-20x^2-47x-15\)

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

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

\(=\left(x-5\right)\left[x^2\left(x+3\right)+3x\left(x+3\right)+\left(x+3\right)\right]\)

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

\(=x^4-5x^3+6x^3-30x^2+10x^2-50x+3x-15\\ =\left(x-5\right)\left(x^3+6x^2+10x+3\right)\\ =\left(x-5\right)\left(x^3+3x^2+3x^2+9x+x+3\right)\\ =\left(x-5\right)\left(x+3\right)\left(x^2+3x+1\right)\)

đặt x^2 - x = t
phương trình trở thành 
t^2-2t-15
= t^2 - 2t + 1 -16
= (t^2 - 1) - 16
=> (x^2 - x - 1) -16

đặt x^2 - x = t
phương trình trở thành 
t^2-2t-15
= t^2 - 2t + 1 -16
= (t^2 - 1) - 16
=> (x^2 - x - 1) -16

(x² - x)² - 2 * (x² - x) - 15 

đặt x² - x = a

có: a² - 2a - 15 = (a² - 2a + 1) - 16 = (a - 1)² - 16 = (a - 5) (a + 3) 

thay vào đc:  (x² - x - 5) (x² - x +3)