$1/_{}$
$a)_{}$ $8x^2y-8xy+2x_{}$
$=2x.(4xy-4y+1)_{}$
$b)_{}$ $x^2-6x-y^2+9_{}$
$=(x-3)^2-y^2_{}$
$=(x-3-y)(x-3+y)_{}$
$c)_{}$ $(x^2+2x)(x^2+4x+3)-24_{}$
$=x.(x+2)(x+1)(x-3)-24_{}$
$=(x^2+3x)(x^2+3x+2)-24_{}$
$=(x^2+3x+1-1)(x^2+3x+1+1)-24_{}$
$=(x^2+3x+1)^2-1-24_{}$
$=(x^2+3x+1)^2-25_{}$
$=(x^2+3x+1-5)(x^2+3x+1+5)_{}$
$=(x^2+3x-4)(x^2+3x+6)_{}$
$2/_{}$
$a)_{}$ $(x+3)^2-(x+2)(x-2)=4x+17_{}$
$⇔x^2+6x+9-(x^2-4)=4x+17_{}$
$⇔x^2+6x+9-x^2+4=4x+17_{}$
$⇔6x+13=4x+17_{}$
$⇔2x=4_{}$
$⇔x=2_{}$
$b)_{}$ $(x-3)(x^2+3x+9)-x.(x^2-4)=1_{}$
$⇔x^3-27-x^3+4x=1_{}$
$⇔-27+4x=1_{}$
$⇔4x=28_{}$
$⇔x_{}=7$
$c)_{}$ $3x^2+7x=10_{}$
$⇔3x^2+7x-10=0_{}$
$⇔3x^2+10x-3x-10=0_{}$
$⇔x.(3x+10)-(3x+10)=0_{}$
$⇔(3x+10)(x-1)=0_{}$
$⇔_{}$ \(\left[ \begin{array}{l}3x+10=0\\x-1=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=-\frac{10}{3}\\x=1\end{array} \right.\)
