Giải
1.
a) $x+y+{{x}^{2}}-{{y}^{2}}$
$=(x+y)+({{x}^{2}}-{{y}^{2}})$
$=(x+y)+(x+y)(x-y)$
$=(x+y)(1+x-y)$

b) ${{x}^{2}}-4{{x}^{2}}{{y}^{2}}+{{y}^{2}}+2xy$
$=({{x}^{2}}+2xy+{{y}^{2}})-4{{x}^{2}}{{y}^{2}}$
$={{\left( x+y \right)}^{2}}-{{\left( 2xy \right)}^{2}}$
$=\left( x+y-2xy \right)\left( x+y+2xy \right)$

c) ${{x}^{3}}-{{x}^{2}}-x+1$
$=({{x}^{3}}-{{x}^{2}})-\left( x-1 \right)$
$={{x}^{2}}\left( x-1 \right)-\left( x-1 \right)$
$=\left( {{x}^{2}}-1 \right)\left( x-1 \right)$
$={{\left( x-1 \right)}^{2}}.\left( x+1 \right)$
d) ${{x}^{3}}+3{{x}^{2}}+3x+1-{{y}^{3}}$
$={{\left( x+1 \right)}^{3}}-{{y}^{3}}$
$=\left( x+1-y \right)\left[ {{\left( x+1 \right)}^{2}}+y\left( x+1 \right)+{{y}^{2}} \right]$
$=\left( x-y+1 \right)\left( {{x}^{2}}+2x+xy+y+{{y}^{2}}+1 \right)$

2.
a) $x\left( x+1 \right)+x+1=0$
$\Leftrightarrow {{x}^{2}}+x+x+1=0$
$\Leftrightarrow {{x}^{2}}+2x+1=0$
$\Leftrightarrow {{(x+1)}^{2}}=0$
$\Leftrightarrow x+1=0$
$\Leftrightarrow x=-1$

b) ${{\left( x+2 \right)}^{2}}-9=0$
$\Leftrightarrow $${{\left( x+2 \right)}^{2}}=9$
$\Leftrightarrow $$x+2=\pm 3$
$\Leftrightarrow $$x\in \{-5;1\}$

c) ${{x}^{3}}+8+\left( x+2 \right)\left( x+7 \right)=0$
$\Leftrightarrow \left( x+2 \right)\left( {{x}^{2}}-2x+4 \right)+\left( x+2 \right)\left( x+7 \right)=0$
$\Leftrightarrow \left( x+2 \right)\left( {{x}^{2}}-2x+4+x+7 \right)=0$
$\Leftrightarrow \left( x+2 \right)\left( {{x}^{2}}-x+11 \right)=0$
$\Leftrightarrow \left( x+2 \right)\left( {{x}^{2}}-2.x.\frac{1}{2}+\frac{1}{4}+11-\frac{1}{4} \right)=0$
$\Leftrightarrow \left( x+2 \right)\left[ {{\left( x-\frac{1}{2} \right)}^{2}}+\frac{43}{4} \right]=0$
$\Leftrightarrow x+2=0$ vì ${{\left( x-\frac{1}{2} \right)}^{2}}+\frac{43}{4} > 0$ với mọi $x$
$\Leftrightarrow x=-2$