Bài 1:
a.
`x^2y - y`
`= y(x^2 - 1)`
`= y(x - 1)(x + 1)`
b.
`x^2 - 25y^2 + 2x + 1`
`=(x^2 + 2x + 1) - (5y)^2`
`=(x + 1)^2 - (5y)^2`
`=(x + 1 - 5y)(x + 1 + 5y)`
c.
`x^3 + 3x^2 - x - 3`
`=(x^3 - x) + (3x^2 - 3)`
`=x(x^2 - 1) + 3(x^2 - 1)`
`=(x^2 - 1)(x + 3)`
`=(x - 1)(x + 1)(x + 3)`
Bài 2:
a.
`x^2 - 5x = 0`
`⇔x(x - 5) = 0`
`⇔`\(\left[ \begin{array}{l}x=0\\x-5=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=5\end{array} \right.\)
Vậy `x = {0; 5}`
b.
`3(x + 2) - x^2 + 4 = 0`
`⇔3(x + 2) - (x^2 - 4) = 0`
`⇔3(x + 2) - (x - 2)(x + 2) = 0`
`⇔(x + 2)(3 - x - 2) = 0`
`⇔(x + 2)(1 - x) = 0`
`⇔`\(\left[ \begin{array}{l}x+2=0\\1-x=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-2\\x=1\end{array} \right.\)
Vậy `x = {-2; 1}`
c.
`x^2 - 5x + 6 = 0`
`⇔x^2 - 3x - 2x + 6 = 0`
`⇔(x^2 - 3x) - (2x - 6) = 0`
`⇔x(x - 3) - 2(x - 3) = 0`
`⇔(x - 3)(x - 2) = 0`
`⇔`\(\left[ \begin{array}{l}x-3=0\\x-2=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=3\\x=2\end{array} \right.\)
Vậy `x = {2; 3}`
