Đáp án:
Bài `1`
`a, x^2 - 7x + 10`
`= x^2 - 2x - 5x + 10`
`= (x^2 - 2x) - (5x - 10)`
`= x(x - 2) - 5(x - 2)`
`= (x - 2)(x - 5)`
`b, x^2 + 6x - 7`
`= x^2 - x + 7x - 7`
`= (x^2 -x) + (7x - 7)`
`= x(x - 1) + 7(x - 1)`
`= (x - 1)(x + 7)`
Bài `2`
`a, (3x - 1)^2 - (x + 2)^2 = 0`
`⇔ [(3x - 1) - (x + 2)](3x - 1 + x + 2) = 0`
`⇔ (3x - 1 - x - 2)(4x + 1) = 0`
`⇔ (2x - 3)(4x + 1) = 0`
`⇒` $\left[\begin{matrix} 2x - 3 = 0\\ 4x + 1 = 0\end{matrix}\right.$ `⇒` $\left[\begin{matrix} 2x = 3\\ 4x = -1\end{matrix}\right.$ `⇒` $\left[\begin{matrix} x = \dfrac{3}{2}\\ x =\dfrac{-1}{4}\end{matrix}\right.$
Vậy `x = 3/2` hoặc `x = -1/4`
`b, x^2 - 6x + 9 = 0`
`⇔ x^2 - 2.x.3 + 3^2 = 0`
`⇔ (x - 3)^2 = 0`
`⇔ x - 3 = 0`
`⇒ x = 3`
Vậy `x = 3`
