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

m . n 3   –   1   +   m   –   n 3     =   ( m n 3   –   n 3 )   +   ( m   - 1 )     =   n 3 ( m   –   1 )   +   ( m   –   1 )     = n 3 + 1 m - 1 =   ( n   +   1 ) ( n 2   –   n   +   1 ) ( m   –   1 )

Đáp án cần chọn là: A

4 x 2   +   4 x   –   y 2   +   1   =   ( ( 2 x ) 2   +   2 . 2 x   +   1 )   –   y 2

=   ( 2 x   +   1 ) 2   –   y 2

= (2x + 1 – y)(2x + 1 + y)

= (2x – y + 1)(2x + y + 1)

Vậy đa thức trong chỗ trống là 2x – y + 1

Đáp án cần chọn là: B

Bài 4: 

Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

n³ -n² -7n +1=n²(n-1) - 7(n-1)-6=0 <=>(n-1).(n²-7)=6

\(A=n^4-n^3-6n^2+7n-21\)

\(A=n^4-3n^3+2n^3-6n^2+7n-21\)

\(A=n^3\left(n-3\right)+2n^2\left(n-3\right)+7\left(n-3\right)\)

\(A=\left(n^3+2n^2+7\right)\left(n-3\right)\)

\(A=n^4-n^3-6n^2+7n-21\)

\(A=n^4-3n^3+2n^3-6n^2+7n-21\)

\(A=n^3\left(n-3\right)+2n^2\left(n-3\right)+7\left(n-3\right)\)

\(A=\left(n^3+2n^2+7\right)\left(n-3\right)\)

a: \(M=m^2\left(m+n\right)-n^2m-n^3\)

\(=m^2\left(m+n\right)-n^2\left(m+n\right)\)

\(=\left(m+n\right)^2\left(m-n\right)\)

\(=\left(-2017+2017\right)^2\cdot\left(-2017-2017\right)\)

=0

b: \(N=n^3-3n^2-n\left(3-n\right)\)

\(=n^2\left(n-3\right)+n\left(n-3\right)\)

\(=n\left(n-3\right)\left(n+1\right)\)

\(=13\cdot10\cdot14=1820\)

\(-8x^3+1=1^3-\left(2x\right)^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)