Lời giải:
\(x^4+6x^2y+9y^2-1=(x^4+3x^2y)+(3x^2y+9y^2)-1\)
\(=x^2(x^2+3y)+3y(x^2+3y)-1\)
\(=(x^2+3y)^2-1=(x^2+3y-1)(x^2+3y+1)\)
\(x^4+6x^2y+9y^2-1\)\(=\left(x^2\right)^2+2.x^2.3y+\left(3y\right)^2-1\)
\(=\left(x^2+3y\right)^2-1\)
\(=\left(x^2+3y-1\right)\left(x^2+3y+1\right)\)