\(1,\\ a,=\left(x-2\right)\left(x-3\right)\\ b,=3\left(x^2+3x-10\right)=3\left(x-2\right)\left(x+5\right)\\ c,=\left(x-1\right)\left(x-2\right)\\ d,=\left(x-6\right)\left(x-3\right)\\ e,=\left(x^3-y^3\right)\left(x^3+y^3\right)=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\\ f,=\left(x+1\right)^3-27z^3=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\\ g,=\left(2x+1\right)^2-9y^2=\left(2x+1-3y\right)\left(2x+1+3y\right)\\ h,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\)
\(2,\\ a,\Rightarrow\left(x+2\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\\ b,\Rightarrow\left(x-8\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=8\end{matrix}\right.\\ c,\Rightarrow\left(x-3\right)\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\\ d,\Rightarrow\left(x+3\right)\left(2x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\\ e,\Rightarrow4x^2-24x+36-4x^2+1=10\\ \Rightarrow-24x=-27\Rightarrow x=\dfrac{9}{8}\\ f,\Rightarrow25x^2+150x+225+1-25x^2=8\\ \Rightarrow150x=-218\Rightarrow x=-\dfrac{109}{75}\)
\(g,\Rightarrow9x^2+18x+9-9x^2+4=10\\ \Rightarrow18x=-3\Rightarrow x=-\dfrac{1}{6}\\ h,\Rightarrow-4x^2+8x-4+4x^2-1=3\\ \Rightarrow8x=8\Rightarrow x=1\\ 3,\\ a,=\left(xyz+xy\right)+\left(xz+yz\right)+\left(x+y\right)+\left(z+1\right)\\ =xy\left(z+1\right)+\left(z+1\right)+z\left(x+y\right)+\left(x+y\right)\\ =\left(xy+1\right)\left(z+1\right)+\left(x+y\right)\left(z+1\right)\\ =\left(xy+1+x+y\right)\left(z+1\right)\\ =\left[x\left(y+1\right)+\left(y+1\right)\right]\left(z+1\right)=\left(x+1\right)\left(y+1\right)\left(z+1\right)\)
\(b,ab\left(a+b\right)-bc\left(b+c\right)-ac\left(c-a\right)\\ =a^2b+ab^2-bc\left(b+c\right)-ac^2+a^2c\\ =\left(a^2b+a^2c\right)+\left(ab^2-ac^2\right)-bc\left(b+c\right)\\ =a^2\left(b+c\right)+a\left(b-c\right)\left(c+b\right)-bc\left(b+c\right)\\ =\left(a^2+ab-ac-bc\right)\left(b+c\right)\\ =\left(a+b\right)\left(a-c\right)\left(b+c\right)\)
