a. \(x^2-2xy+y^2-z^2=\\ \left(x^2-2xy+y^2\right)-z^2\\ =\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
b. \(x^2-6x+9-9y\\ =\left(x^2-6x\right)+\left(9-9y\right)\\ =x\left(x-6\right)+9\left(1-y\right)\)
1,
a, \(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2\)
=\(\left(x-y-z\right)\left(x-y+z\right)\)
b, hình như sai đề phải ko bn?
Phải là: \(x^2-6x+9-9y^2\)=\(\left(x-3\right)^2-\left(3y\right)^2\)\(=\left(x-3y-3\right)(x+3y-3)\)
c,\(\left(x+y\right)\left(y+z\right)\left(x+z\right)+xyz\)
=\((x^2+xy+yz+xz)\left(z+y\right)+xyz\)
=\((x^2z+xyz+xz^2)+(x^2y+xy^2+xyz)+\)\(\left(yz^2+y^2z+xyz\right)\)
= xz(x + y + z) + xy(x + y + z) + yz(x + y +z)
=(x+y+z)(xz+xy+yz)
2,a,\(\left(x-2\right)\left(x+1\right)=0\)\(\Rightarrow\)\(\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b,Đề phải thế này nha:
\(5x\left(x-3\right)-x+3=0\)\(\Rightarrow\)(x - 3)(5x - 1)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\)\(\Rightarrow\)\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
Chúc các bn học tốt 
Bài 2
a]x.[x-2]+x-2=0-> (x+1).(x-2)=0
-> (x+1)=0;(X-2)=0
->x=-1;x=2
a. \(x\left(x-3\right)+x-2=0\\ \Leftrightarrow x^2-3x+x-2=0\\ \Leftrightarrow x^2-2x=2\\ \Leftrightarrow x\left(x-2\right)=2\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\x-2=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
b \(5x\left(x-3-x+3\right)=0\\ \Leftrightarrow5x=0\\ \Leftrightarrow x=0\)
