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

\(ab\left(x^2+y^2\right)+xy\left(a^2+b^2\right)=abx^2+aby^2+a^2xy+b^2xy\)

\(=ax\left(bx+ay\right)+by\left(ay+bx\right)=\left(ax+by\right)\left(ay+bx\right)\)

TL:

\(A=9x^2-y^2+6x+1\)

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

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

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

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

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

\(2xy-x^2-y^2+16\)

\(=16-\left(x-y\right)^2\)

\(=\left(4-x+y\right)\left(4+x-y\right)\)

Mấy bài này khá đơn giản .

Bạn chỉ cần áp dụng hằng đẳng thức \(x^2-y^2=\left(x+y\right)\left(x-y\right)\) là được nhs =))

a)

\(\left(xy+4\right)^2-4\left(x+y\right)^2\)

\(=\left(xy+4\right)^2-\left[2\left(x+y\right)\right]^2\)

\(=\left[xy+4-2\left(x+y\right)\right]\left[xy+4+2\left(x+y\right)\right]\)

\(=\left(xy+4-2x-2y\right)\left(xy+4+2x+2y\right)\)

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

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

b)

\(\left(ab-xy\right)^2-\left(bx-ay\right)^2\)

\(=\left(ab-xy-bx+ay\right)\left(ab-xy+bx-ay\right)\)

\(=\left[a\left(b+y\right)-x\left(b+y\right)\right]\left[a\left(b-y\right)+x\left(b-y\right)\right]\)

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

c)

\(=\left(x^2+8x-34+3x^2-8x-2\right)\left(x^2+8x-34-3x^2+8x+2\right)\)

\(=\left(4x^2-36\right)\left(-2x^2+16x-32\right)\)

\(=\left(2x-6\right)\left(2x+6\right)\left(-2\right)\left(x^2-8x+16\right)\)

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

Bạn liểm tra lại nhs

Mk lm hay nhấm lắm

=))

Câu a ảo vc

b,<=>xya² + xyb² - abx² - aby²

   <=>xa ( ya - bx ) - by ( ay - bx )

   <=>( xa - bx )( ya - bx )

c, <=>( 5x - 1 )² - ( 3y )²

    <=>( 5x - 1 - 3y ) ( 5x - 1 + 3y) 

\(ab\left(x^2+y^2\right)-xy\left(a^2+b^2\right)\)

\(=abx^2+aby^2-a^2xy-b^2xy\)

\(=\left(abx^2-b^2xy\right)-\left(a^2xy-aby^2\right)\)

\(=bx\left(ax-by\right)-ay\left(ax-by\right)\)

\(=\left(ax-by\right)\left(bx-ay\right)\)

a/ (x⁴ - 6x² + 9) - x² = (x² - 3)² - x² = (x² - 3 - x)(x² - 3 + x)

a/ (x² - 3-x)(x2 ​- 3+x)

b/ (ay - bx)(by - ax)