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

=ax^3 -xy^3 - ax^3 -x^3y +a^3x -a^3y

=-xy^3 - x^3y

k mk đi

ai k mk 

mk k lại

thanks

  ( a - x )y³ - ( a - y )x³  + ( x - y )a³

= ay³  + a²y²  -  ax²y  -  a²xy  - a²y²  -  a³y + a²x² + a³x - xy³ - axy² + x³y + ax²y + axy² + a²xy - ax³ - a²x²

= ay( y² +ay -x² - ax ) - a²( y² + ay -x² -ax ) - xy( y² + ay - x² -ax ) + ax( y² + ay -x² -ax )

= ( y² + ay - x² - ax )( ay - a² - xy + ax )

= ( y² + xy +ay -xy -ax -x² )[ ( y -a )a - x( y-a ) ]

= [ y( y +x +a ) - x( y + x + a )]( a - x )( a - y)

= ( y + x + a)( y -x )( a - x)( y - a)

#)Bạn tham khảo nhé :

Câu hỏi của Linh Mà - Toán lớp 9 | Học trực tuyến

P/s : vô thống kê hỏi đáp của mk có thể dùng đc link nhé

Tham khảo tại đây:

https://h.vn/hoi-dap/question/194l358.htm

b: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-3\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+21\)

\(=\left(x^2+5x+3\right)\left(x^2+5x+7\right)\)

a, \(\left(x+y\right)^3+\left(x-y\right)^3=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

\(=2x\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)=2x\left(x^2+3y^2\right)\)

b, \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-3\)

Đặt \(x^2+5x+5=t\)

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

Theo cách đặt \(\left(x^2+5x+3\right)\left(x^2+5x+7\right)\)

a)\(\left(x+y\right)^3+\left(x-y\right)^3\)
\(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=2x\left(x^2+2xy+y^2-x^2+xy-xy+y^2+x^2-2xy+y^2\right)\)
\(=2x\left(x^2+3y^2\right)\)

t chỉ cho kết quả thôi nhá, còn nhóm nhân tử you tự xử nhá !

=(x-y)(z-x)(z-y)(x+y+z)

\(\left(x-y\right)z^3+\left(z-z\right)y^3+\left(y-z\right)x^3\)

\(=z^3\left(x-y\right)+y^3\left(z-x\right)+x^3\left(y-z\right)\)

\(=xz^3-yz^3+\left(z-x\right)y^3+\left(y-z\right)x^3\)

\(=xz^3-yz^3+y^3z-xy^3+\left(y-z\right)x^3\)

\(=xz^3-yz^3+y^3z-xy^3+y^3z-xy^3+x^3y-x^3z\)

Mk ko chắc

\(\left(x-y\right)z^3+\left(z-x\right)y^3+\left(y-z\right)x^3\)

\(=\left(x-y\right)z^3-\left[\left(x-y\right)+\left(y-z\right)\right]y^3+\left(y-z\right)x^3\)

\(=\left(x-y\right)z^3-\left(x-y\right)y^3-\left(y-z\right)y^3+\left(y-z\right)x^3\)

\(=\left(x-y\right)\left(z^3-y^3\right)+\left(y-z\right)\left(x^3-y^3\right)\)

\(=\left(x-y\right)\left(z-y\right)\left(z^2+zy+y^2\right)+\left(y-z\right)\left(x-y\right)\left(x^2+y^2+xy\right)\)

\(=\left(x-y\right)\left(y-z\right)\left(x^2+y^2+xy-z^2-y^2-zy\right)\)

\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\left(x+y+z\right)\)

\(a,\)

\(x^3y-y\)

\(=y\left(x^3-1\right)\)

\(=y\left[\left(x-1\right)\left(x^2+x+1\right)\right]\)

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

\(b,\)

\(x^3y+y\)

\(=y\left(x^3+1\right)\)

\(=y\left[\left(x+1\right)\left(x^2-x+1\right)\right]\)

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

\(c,\)

\(\left(x-y\right)^2-x\left(y-x\right)\)

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

\(=\left(x-y\right)\left(x-y+x\right)\)

\(=\left(x-y\right)\left(2x-y\right)\)