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

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

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

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

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

Đặt \(x^2+x+1=t\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=t\left(t+1\right)-12=t^2+t-12=\left(t^2+t+\dfrac{1}{4}\right)-\dfrac{49}{4}=\left(t+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(t+\dfrac{1}{2}-\dfrac{7}{2}\right)\left(t+\dfrac{1}{2}+\dfrac{7}{2}\right)=\left(t-3\right)\left(t+4\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

= \(\left(x^2+x+1\right)\left[\left(x^2+x+1\right)+1\right]-12\)

= \(\left(x^2+x+1\right)^2\left(x^2+x+1\right)-12\)

= \(\left(x^2+x+1\right)\left(x^2+x+1\right)-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-4.3\)

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

= \(\left(x^2+x+5\right)\left(x^2+x-2\right)\)

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

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

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

=3xy(x+y)+3(x+y)^2z+3(x+y)z^2

=3(x+y)(xy+(x+y)z+z^2)

=3(x+y)(xy+xz+yz+z^2)

=3(x+y)(x(y+z)+z(y+z))

=3(x+y)(y+z)(x+z)

1) a^3-3a^2b+3ab^2-b^3+b^3-3b^2c+3bc^2-c^3+c^3-3c^2a+3ca^2-a^3

= -3(a^2b-ab^2+b^2c-bc^2+c^2a-ca^2)

=-3(ab(a-b)+c(b^2-a^2)-c^2(b-a))

= -3(ab(a-b)-c(a+b)(a-b)+c^2(a-b))

= -3(a-b)(ab-ac-bc+c^2)

= -3(a-b)(a(b-c)-c(b-c))

= -3(a-b)(b-c)(a-c)

\(x^2-y^2=\left(x-y\right)\left(x+y\right)\)\(=>\)Hằng đẳng thức Hiệu hai bình phương 

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

Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)

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

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

\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)

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

\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)

\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

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

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

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

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

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

P/s: Tham khảo nha!!

a

Ta có

\(2x^2+2x=2x\left(x+1\right)\)

b

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

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