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

`a)3x-3a+yx-ya`

`=3(x-a)+y(x-a)`

`=(x-a)(y+3)`

`b)x^2-9-4(x+3)`

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

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

`=(x+3)(x-7)`

a)  \(=3x+yx-\left(3a+ya\right)\) \(=x\left(3+y\right)-a\left(3+y\right)\) \(=\left(3+y\right)\left(x-a\right)\)

b) \(=\left(x-3\right)\left(x+3\right)-4\left(x+3\right)\)  \(=\left(x+3\right)\left(x-3-4\right)\) \(=\left(x+3\right)\left(x-7\right)\)

(2 - 3a)² - (3 - a)²

= (2 - 3a - 3 + a)(2 - 3a + 3 - a)

= (-2a - 1)(1 - 4a)

Ta có:A= 3a+3b-a^2-ab

=>A= (3a-a^2)+(3b-ab)

=>A= a(3-a)+b(3-a)

=>A= (a+b)(3-a)

tớ mới học lớp 6

mk mới hk lớp 6 thui ak

a²-b²+3a+3b=(a-b)(a+b)+3(a+b)=(a+b)(a-b+3)

=(a-b).(a+b)+3.(a+b)=(a+b).(a-b+3)

Tưởng tượng nhé :v
(a+b)^2 + 3a+b)+10= t^2 +3t+10 ( đặ a+b=1)  = (t^2+3t+9/4) +31/4 >0  
=> Không thể phân tích :3