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

\(\Rightarrow\)\(\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\)\(\Rightarrow\)\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

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