a: \(=x^2\left(x-2\right)\)

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

c: =(x-3)(x+2)

Lời giải:
$=x+\sqrt{x}(\sqrt{y}+\sqrt{2})-\sqrt{3}(\sqrt{y}+\sqrt{2})-3$

$=(x-3)+\sqrt{x}(\sqrt{y}+\sqrt{2})-\sqrt{3}(\sqrt{y}+\sqrt{2})$

$=(\sqrt{x}-\sqrt{3})(\sqrt{x}+\sqrt{3})+(\sqrt{y}+\sqrt{2})(\sqrt{x}-\sqrt{3})$

$=(\sqrt{x}-\sqrt{3})(\sqrt{x}+\sqrt{3}+\sqrt{y}+\sqrt{2})$

\(x-y-\sqrt{x}-\sqrt{y}\\ =x-y-\left(\sqrt{x}+\sqrt{y}\right)\\ =\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}+\sqrt{y}\right)\\ =\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}-1\right)\)

=(x-y)-(căn x+căn y)

=(căn x-căn y)(căn x+căn y)-(căn x+căn y)

=(căn x+căn y)(căn x-căn y-1)

\(=\left(x\sqrt{x}+y\sqrt{y}\right)+\left(x-y\right)\)

\(=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}+\sqrt{y}\right)\left(x+y-\sqrt{xy}+\sqrt{x}-\sqrt{y}\right)\)

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

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

= x(x^2 - 1) + y(x^2 - 1)

= ( x^2 -1)(x+y)

\(x^2-9x-y^2-9y\)

\(=\left(x^2-y^2\right)-\left(9x+9y\right)\)

\(=\left(x-y\right)\left(x+y\right)-9\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-9\right)\)

1. 

\(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\\ =\left(12x^2+6x\right)\left(y+z+y-z\right)\\ =2y\left(12x^2+6x\right)\\ =2y.6x\left(2x+1\right)\\ =12xy\left(2x+1\right)\)

2. 

\(x\left(x-6\right)+10\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)

Vậy \(x\in\left\{6;-10\right\}\) là nghiệm của pt

Bài 1:

Ta có: \(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\)

\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)

\(=6x\left(2x+1\right)\cdot2y\)

\(=12xy\left(2x+1\right)\)

Bài 2: 

Ta có: \(x\left(x-6\right)+10\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)