\(h,x^2-6x-y^2+9\\ =\left(x^2-6x+9\right)-y^2\\ =\left(x-3\right)^2-y^2\\ =\left(x-y-3\right)\left(x+y-3\right)\\ j,x^3-8\\ =x^3-2^3\\ =\left(x-2\right)\left(x^2+2x+4\right)\\ k,4x^4+20x^2+25\\ =\left(2x^2\right)^2+2\cdot2x^2\cdot5+5^2\\ =\left(2x^2+5\right)^2\\ l,x^2+1-16y^2+2x\\ =\left(x^2+2x+1\right)-\left(4y\right)^2\\ =\left(x+1\right)^2-\left(4y\right)^2\\ =\left(x-4y+1\right)\left(x+4y+1\right)\\ m,x^2-y^2-3x+3y\\ =\left(x+y\right)\left(x-y\right)-3\left(x-y\right)\\ =\left(x-y\right)\left(x+y-3\right)\\ n,3a^2-6ab+3b^2-12c^2\\ =3\left[\left(a^2-2ab+b^2\right)-4c^2\right]\\ =3\left[\left(a-b\right)^2-\left(2c\right)^2\right]\\ =3\left(a-b-2c\right)\left(a-b+2c\right)\)
H)
= \[
x^2 - 6x + 9 - y^2 = 0
\]
= \[
(x - 3)^2 - y^2 = 0
\]
= \[
(a^2 - b^2) = (a - b)(a + b)
\]
= \[
(x - 3 - y)(x - 3 + y)
\]
J)
= \[
x^3 - 2^3 = (x - 2)(x^2 + 2x + 4)
\]
K)
\[
= 4x^4 + 20x^2 + 25 = (2x^2)^2 + 2(2x^2)(5) + 5^2
\]
= \[
(2x^2 + 5)^2
\]
L)
= \[
x^2 + 2x + 1 - 16y^2 = 0
\]
= \[
(x + 1)^2 - 16y^2 = 0
\]
= \[
(x + 1 - 4y)(x + 1 + 4y)
\]
M)
= \[
x^2 - 3x - y^2 + 3y = 0
\]
= \[
(x^2 - 3x) - (y^2 - 3y) = 0
\]
= \[
x^2 - 3x = (x - \frac{3}{2})^2 - \frac{9}{4} \quad \text{và} \quad y^2 - 3y = (y - \frac{3}{2})^2 - \frac{9}{4}
\]
= \[
((x - \frac{3}{2})^2 - \frac{9}{4}) - ((y - \frac{3}{2})^2 - \frac{9}{4}) = 0
\]
= \[
(x - \frac{3}{2} - (y - \frac{3}{2}))(x - \frac{3}{2} + (y - \frac{3}{2}))
\]
N)
= \[
3(a^2 - 2ab + b^2) - 12c^2
\]
= \[
3((a - b)^2 - 4c^2) = 0
\]
= \[
3(a - b - 2c)(a - b + 2c)
\]
