Lời giải:
$x^2-3xy+x-3y=(x^2-3xy)+(x-3y)=x(x-3y)+(x-3y)=(x-3y)(x+1)$
$x^2+6x-y^2+9=(x^2+6x+9)-y^2=(x+3)^2-y^2=(x+3-y)(x+3+y)$
$x^3-9x+3x^2-27=(x^3+3x^2)-(9x+27)$
$=x^2(x+3)-9(x+3)=(x^2-9)(x+3)=(x-3)(x+3)(x+3)=(x-3)(x+3)^2$
$7x^2-7xy-4x+4y=(7x^2-7xy)-(4x-4y)=7x(x-y)-4(x-y)=(7x-4)(x-y)$
$2x^2+4ax+x+2a=(2x^2+4ax)+(x+2a)=2x(x+2a)+(x+2a)=(x+2a)(2x+1)$
$x^4-9+3x^3-9x=(x^4-3^2)+(3x^3-9x)=(x^2-3)(x^2+3)+3x(x^2-3)$
$=(x^2-3)(x^2+3+3x)$
\(x^2-3xy+x-3y\)
\(=\left(x^2-3xy\right)+\left(x-3y\right)\)
\(=x\left(x-3y\right)+\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+1\right)\)
\(---\)
\(x^2+6x-y^2+9\)
\(=\left(x^2+6x+9\right)-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3-y\right)\left(x+3+y\right)\)
\(---\)
\(x^3-9x+3x^2-27\)
\(=\left(x^3+3x^2\right)-\left(9x+27\right)\)
\(=x^2\left(x+3\right)-9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-9\right)\)
\(=\left(x+3\right)^2\left(x-3\right)\)
\(---\)
\(7x^2-7xy-4x+4y\)
\(=\left(7x^2-7xy\right)-\left(4x-4y\right)\)
\(=7x\left(x-y\right)-4\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-4\right)\)
\(---\)
\(2x^2+4ax+x+2a\)
\(=\left(2x^2+4ax\right)+\left(x+2a\right)\)
\(=2x\left(x+2a\right)+\left(x+2a\right)\)
\(=\left(x+2a\right)\left(2x+1\right)\)
\(---\)
\(x^4-9+3x^3-9x\)
\(=\left(x^4-9\right)+\left(3x^3-9x\right)\)
\(=\left(x^2-3\right)\left(x^2+3\right)+3x\left(x^2-3\right)\)
\(=\left(x^2-3\right)\left(x^2+3+33x\right)\)
#\(Toru\)