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

a) Ta có: \(\dfrac{19}{33}=\dfrac{38}{66};\dfrac{6}{12}=\dfrac{1}{2}=\dfrac{33}{66};\dfrac{13}{22}=\dfrac{39}{66}\)

Mà \(\dfrac{33}{66}< \dfrac{38}{66}< \dfrac{39}{66}\Rightarrow\dfrac{6}{12}< \dfrac{19}{33}< \dfrac{13}{22}\)

Vậy các số hữu tỉ được sắp xếp theo thứ tự tăng dần là: \(\dfrac{6}{12};\dfrac{19}{33};\dfrac{13}{22}\)

b) Ta có:

\(\dfrac{-18}{12}=\dfrac{-3}{2}=\dfrac{-105}{70};\dfrac{-10}{7}=\dfrac{-100}{70};\dfrac{-8}{5}=\dfrac{-112}{70}\)

Mà \(\dfrac{-112}{70}< \dfrac{-105}{70}< \dfrac{-100}{70}\Rightarrow\dfrac{-8}{5}< \dfrac{-18}{12}< \dfrac{-10}{7}\)

Vậy các số hữu tỉ được sắp xếp theo thứ tự tăng dần là: \(\dfrac{-8}{5};\dfrac{-18}{12};\dfrac{-10}{7}\)

\(a)\left(3+xy^2\right)^2=3^2+2.3.xy^2+\left(xy^2\right)^2=9+6xy^2+x^2y^4\\ b)\left(10-2m^2n\right)^2=10^2-2.10.2m^2n+\left(2m^2n\right)^2=100-40m^2n+4m^4n^2\\ c)\left(a-b^2\right)\left(a+b^2\right)=a^2-\left(b^2\right)^2=a^2-b^4\)

\(x\left(2x-1\right)+\dfrac{1}{3}-\dfrac{2}{3}x=0\\ \Leftrightarrow x\left(2x-1\right)-\dfrac{1}{3}\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x-\dfrac{1}{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x-\dfrac{1}{3}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(g)2bx-3ay-6by+ax\\ =\left(2bx-6by\right)+\left(ax-3ay\right)\\ =2b\left(x-3y\right)+a\left(x-3y\right)\\ =\left(x-3y\right)\left(a+2b\right)\\ h)x+2a\left(x-y\right)-y\\ =2a\left(x-y\right)+\left(x-y\right)\\ =\left(x-y\right)\left(2a+1\right)\\ e)a^5-a^3+a^2-1\\ =a^3\left(a^2-1\right)+\left(a^2-1\right)\\ =\left(a^2-1\right)\left(a^3+1\right)\\ =\left(a-1\right)\left(a+1\right)^2\left(a^2-a+1\right)\)

\(1)x^3-2x^2+x\\ =x\left(x^2-2x+1\right)\\ =x\left(x-1\right)^2\\ 2)2x^2+4x+2-2y^2\\ =2\left(x^2+2x+1-y^2\right)\\ =2\left[\left(x+1\right)^2-y^2\right]\\ =2\left(x-y+1\right)\left(x+y+1\right)\\ 3)2xy-x^2-y^2+16\\ =16-\left(x^2-2xy+y^2\right)\\ =16-\left(x-y\right)^2=\left(x-y+4\right)\left(y-x+4\right)\\ 4)x^3+2x^2y+xy^2-9x\\ =x\left(x^2+2xy+y^2-9\right)\\ =x\left[\left(x+y\right)^2-9\right]\\ =x\left(x+y-3\right)\left(x+y+3\right)\\ 5)2x-2y-x^2+2xy-y^2\\ =\left(2x-2y\right)-\left(x^2-2xy+y^2\right)\\ =2\left(x-y\right)-\left(x-y\right)^2\\ =\left(x-y\right)\left(y-x+2\right)\)

\(1.\\ a)x^8+x^4+1\\ =\left(x^8+2x^4+1\right)-x^4\\ =\left(x^4+1\right)^2-x^4\\ =\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\)

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

\(-36+24x-x^2\\ =108-\left(x^2-24x+144\right)\\ =\left(6\sqrt{3}\right)^2-\left(x-12\right)^2\\ =\left(x-12+6\sqrt{3}\right)\left(12-x+6\sqrt{3}\right)\)

\(\left(2x+3\right)^2-\left(x-1\right)^2=0\\ \Leftrightarrow\left(2x+3-x+1\right)\left(2x+3+x-1\right)=0\\ \Leftrightarrow\left(x+4\right)\left(3x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{-2}{3}\end{matrix}\right.\)

\(2016x^2-x-2017=0\\ \Leftrightarrow2016x^2+2016x-2017x-2017=0\\ \Leftrightarrow2016x\left(x+1\right)-2017\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(2016x-2017\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\2016x-2017=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{2017}{2016}\end{matrix}\right.\)