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

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

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

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

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

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

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

a) ( x + 1 )( x + 2 )( x + 3 )( x + 4 ) - 15

= [ ( x + 1 )( x + 4 ) ][ ( x + 2 )( x + 3 ) ] - 15

= ( x² + 5x + 4 )( x² + 5x + 6 ) - 15 (*)

Đặt t = x² + 5x + 4 

(*) trở thành

t( t + 2 ) - 15

= t² + 2t - 15

= t² - 3t + 5t - 15

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

= ( t - 3 )( t + 5 )

= ( x² + 5x + 4 - 3 )( x² + 5x + 4 + 5 )

= ( x² + 5x + 1 )( x² + 5x + 9 )

b) ( x + 2 )( x + 3 )²( x + 4 ) - 12

= [ ( x + 2 )( x + 4 ) ]( x + 3 )² - 12

= ( x² + 6x + 8 )( x² + 6x + 9 ) - 12 (*)

Đặt t = x² + 6x + 8

(*) trở thành

t( t + 1 ) - 12

= t² + t - 12

= t² - 3t + 4t - 12

= t( t - 3 ) + 4( t - 3 )

= ( t - 3 )( t + 4 )

= ( x² + 6x + 8 - 3 )( x² + 6x + 8 + 4 )

= ( x² + 6x + 5 )( x² + 6x + 12 )

= ( x² + x + 5x + 5 )( x² + 6x + 12 )

= [ x( x + 1 ) + 5( x + 1 ) ]( x² + 6x + 12 )

= ( x + 1 )( x + 5 )( x² + 6x + 12 )

a, Gọi\(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)

                \(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)

Đặt\(y=x^2+5x+4\)

\(\Rightarrow A=y\left(y+2\right)-15\)

        \(=y^2+2y-15\)

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

Hay\(A=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

Vậy...

b,Gọi\(B=\left(x+2\right)\left(x+3\right)^2\left(x+4\right)-12\)

           \(=\left(x^2+6x+8\right)\left(x^2+6x+9\right)-12\)

Đặt\(z=x^2+6x+8\)

\(\Rightarrow B=z\left(z+1\right)-12\)

        \(=z^2+z-12\)

        \(=\left(z-3\right)\left(z+4\right)\)

Hay\(B=\left(x^2+6x+5\right)\left(x^2+6x+12\right)\)

Vậy...

Linz

\(a,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-15\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)(*)

Đặt \(t=x^2+5x+4\)

(*)\(=t\left(t+2\right)-15\)

\(=t^2+2t-15\)

\(=t^2-3t+5t-15\)

\(=t\left(t-3\right)+5\left(t-3\right)\)

\(=\left(t-3\right)\left(t+5\right)\)

\(=\left(x^2+5x+4-3\right)\left(x^2+5x+4+3\right)\)

\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

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

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

\(=\left(x^2+6x+8\right)\left(x^2+6x+9\right)-12\)(*)

Đặt \(t=x^2+6x+8\)

(*)\(=t\left(t+1\right)-12\)

\(=t^2+t-12\)

\(=t^2-3t+4t-12\)

\(=t\left(t-3\right)+4\left(t-3\right)\)

\(=\left(t+4\right)\left(t-3\right)\)

\(=\left(x^2+6x+12\right)\left(x^2+6x+5\right)\)

\(=\left(x^2+6x+12\right)\left(x^2+x+5x+5\right)\)

\(=\left(x^2+6x+12\right)\left[x\left(x+1\right)+5\left(x+1\right)\right]\)

\(=\left(x^2+6x+12\right)\left(x+5\right)\left(x+1\right)\)

\(4x^3-7x^2+3x\)

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

\(=4x^2\left(x-1\right)-3x\left(x-1\right)\)

\(=\left(x-1\right)\left(4x^2-3x\right)=x\left(x-1\right)\left(4x-3\right)\)

\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-15\)

\(=\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)-15\)

\(=\left(x^2-5x+4\right)\left(x^2-5x+4+2\right)-15\)

\(=\left(x^2-5x+4\right)^2+2\left(x^2-5x+4\right)+1-16\)

\(=\left(x^2-5x+4+1\right)^2-4^2\)

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

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

mở sách giải ra mà cop

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

a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-15\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)(1)

Đặt \(x^2+5x+4=t\)

\(\Rightarrow\left(1\right)=t\left(t+2\right)-15=t^2+2t+1-16\)

\(=\left(t+1\right)^2-4^2=\left(t+5\right)\left(t-3\right)\)

\(=\left(x^2+5x+9\right)\left(x^2+5x+1\right)\)

b) \(\left(2x+5\right)^2-\left(x-9\right)^2\)

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

\(=\left(3x-4\right)\left(x+14\right)\)

a) (x+1)(x+2)(x+3)(x+4) -15

= (x+1)(x+4)(x+2)(x+3)-15

=(x²+5x+4)(x²+5x+6)-15 (*)

đặt x²+5x+5 = k ( k khác 0 )

thì (*) = (k-1)(k+1)-15

=k²-1-15=k²-16

= (k+4)(k-4)

=(x²+5x+9)(x²+5x+1)

b) (2x+5)²-(x-9)²

=(2x+5+x-9)(2x+5-x+9)

=(3x-4)(x+14)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)