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

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

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

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

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

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

bạn kiểm tra lại nhé :)

(x+y)^3-1-3(x+y)(x+y-1)=(x+y)^3-(3x+3y)(x+y-1)-1

                                     =(x+y)^3-(3x^2+6xy+3y^2-3x-3y)-1

                                     =(x+y)^3-3(x^2+2xy+y^2)-3(x+y)-1

                                     =(x+y)^3-3(x+y)^2-3(x+y)-1

xong đặt nhân tử rút ra, rồi dùng HĐT a^2-1 nha

máy hết pin rùi

mk lộn thứ tự phải là câu 1->3->2->4

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

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

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

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

 = y^2-y+x^2-3x+2

 = (x^2-3x+2,25)-(y^2+y+0,25)

 = (x-3/2)^2 - (y+1/2)^2

 = (x-3/2+y+1/2).(x-3.2-y-1/2)

 =(x+y-1).(x-y-2)

k mk nha

\(2xy+\frac{1}{4}-x^2-y^2\)\(=\left(\frac{1}{2}\right)^2-\left(x^2-2xy+y^2\right)\)\(=\left(\frac{1}{2}\right)^2-\left(x-y\right)^2=\left(\frac{1}{2}+x-y\right)\left(\frac{1}{2}-x+y\right)\)

2xy + 1/4 - x² - y²

=> - (-2xy + x² + y² ) + 1/4

=> - (x - y)² + (1/2)²

=> - (x - y + 1/2)(x - y - 1/2)

ĐÚNG ĐÓ NHA, NHỚ NHÉ!

     y^2 - ( x^2 - 2x + 1 )

⇔ y^2 - ( x - 1 )^2

⇔ ( y - x - 1 ) ( y + x - 1 )

nha bạn 

  ( x + y )² - 2( x + y ) + 1 = 

= ( x + y - 1 )² ( áp dụng hằng đẳng thức thứ bình phương 1 hiệu nha )

Hok tốt!!!!!!!!!!!!!!!!