$a)_{}$ $x^2+2x+1-4_{}$
$=x^2+2x-3_{}$
$=x^2+3x-x-3_{}$
$=(x^2+3x)-(x+3)_{}$
$=x.(x+3)-(x+3)_{}$
$=(x-1)(x+3)_{}$
$b)_{}$ $x^2+2xy+y^2-4_{}$
$=(x^2+2xy+y^2)-4_{}$
$=(x+y)^2-2^2_{}$
$=(x+y-2)(x+y+2)_{}$
$c)_{}$ $x^2y+xy^2+2xy_{}$
$=xy.(x+y+2)_{}$
$Tìm_{}$ $x:_{}$
$a)_{}$ $x^2-4x+4=0_{}$
$⇔(x-2)^2=0_{}$
$⇔x-2=0_{}$
$⇔x=2_{}$
$Vậy_{}$ $x=2_{}$
$b)_{}$ $x^3-4x^2+3x-12=0_{}$
$⇔(x^3-4x^2)+(3x-12)=0_{}$
$⇔x^2.(x-4)+3(x-4)=0_{}$
$⇔(x^2+3)(x-4)=0_{}$
$⇔_{}$ \(\left[ \begin{array}{l}x^2+3=0\\x-4=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x∉R\\x=4\end{array} \right.\)
$Vậy_{}$ $x=4_{}$
