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

`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)

a) Ta có: \(a^2-b^2-5a+5b\)

\(=\left(a-b\right)\left(a+b\right)-5\left(a-b\right)\)

\(=\left(a-b\right)\left(a+b-5\right)\)

b) Ta có: \(a^2-b^2-3ab^2-3a^2b\)

\(=\left(a-b\right)\left(a+b\right)-3ab\left(a+b\right)\)

\(=\left(a+b\right)\left(a-b-3ab\right)\)

c) Ta có: \(x^2-xy+2y-4\)

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

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

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

d) Ta có: \(4a^2-10ax+15x-9\)

\(=\left(4a^2-9\right)-\left(10ax-15x\right)\)

\(=\left(2a-3\right)\left(2a+3\right)-5x\left(2a-3\right)\)

\(=\left(2a-3\right)\left(2a+3-5x\right)\)

a,\(5ab-45a^3b\)

=\(5ab\left(1-9a^2\right)\)

=\(5ab\left(1-3a\right)\left(1+3a\right)\)

b,\(3a-6ab+5-10b\)

=\(\left(3a-6ab\right)+\left(5-10b\right)\)

=\(3a\left(1-2b\right)+5\left(1-2b\right)\)

=\(\left(1-2b\right)\left(3a+5\right)\)

c,\(a^2-7ab-2a+14b\)

=\(\left(a^2-7ab\right)-\left(2a-14b\right)\)

=\(a\left(a-7b\right)-2\left(a-7b\right)\)

=\(\left(a-7b\right)\left(a-2\right)\)

d,\(4a^2-8b+4a-8ab\)

=\(\left(4a^2-8ab\right)+\left(4a-8b\right)\)

=\(4a\left(a-2b\right)+4\left(a-2b\right)\)

=\(\left(a-2b\right)\left(4a+4\right)\)

=\(4\left(a-2b\right)\left(a+1\right)\)

e,\(a^2-5a+15b-9b^2\)

=\(\left(a^2-9b^2\right)-\left(5a-15b\right)\)

=\(\left(a-3b\right)\left(a+3b\right)-5\left(a-3b\right)\)

=\(\left(a-3b\right)\left(a+3b-5\right)\)

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

(4x-y)² - (3a)²

=(4x-y+3a)(4x-y+3a)

Học tốt!!!!!!!!!!!!!!!!!

ღ๖ۣۜLinh's ๖ۣۜLinh'sღ] ★we are one★ hai dấu cộng à bn

nhầm nhâm

=(4x-y+3a)(4x-y-3a)

Sorry nhiều nhá

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)