Giúp mình với cảm ơn

$\text{ @HV }$

$\textit{ BÀI 1: }$

` a) x^2 + xy - 6x - 6y `

` = x. (x + y) - 6. (x + y) `

` = (x - 6). (x + y) `

` b) x^2y - 8xy + 16y `

` = x^2y - 4xy - 4xy + 16y `

` = xy. (x - 4) - 4y. (x - 4) `

` = (xy - 4y). (x - 4) `

` = y. (x - 4). (x - 4) `

` = y. (x - 4)^2 `

` c) x^2 - 9x + 8 `

` = x^2 - x - 8x + 8 `

` = x. (x - 1) - 8. (x - 1) `

` = (x - 8). (x - 1) `

` d) 4x^2 - 9 + (2x - 3). (x - 5) `

` = (2x - 3). (2x + 3) + (2x - 3). (x - 5) `

` = (2x - 3). (2x + 3 + x - 5) `

` = (2x - 3). (3x - 2) `

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$\textit{ BÀI 2: }$

` a) 3x. (2 - x) = 0 `

⇔ $\left[\begin{matrix} 3x = 0\\ 2 - x = 0\end{matrix}\right.$

⇔ $\left[\begin{matrix} x = 0\\ x = 2\end{matrix}\right.$

Vậy ` x = 0 ` và ` x = 2 `

` b) x. (2 - 5x) - (5x - 2) = 0 `

` ⇔ x. (2 - 5x) + (2 - 5x) = 0 `

` ⇔ (x + 1). (2 - 5x) = 0 `

⇔ $\left[\begin{matrix} x + 1 = 0\\ 2 - 5x = 0\end{matrix}\right.$

⇔ $\left[\begin{matrix} x = -1\\ x = \dfrac{2}{5}\end{matrix}\right.$

Vậy ` x = -1 ` và ` x = \frac{2}{5} `

` c) (2x + 1)^3 - 9. (2x + 1) = 0 `

` ⇔ [(2x + 1)^2 - 9]. (2x + 1) = 0 `

` ⇔ (2x + 1 + 3). (2x + 1 - 3). (2x + 1) = 0 `

` ⇔ (2x + 4). (2x - 2). (2x + 1) = 0 `

⇔ $\left[\begin{matrix} 2x + 4 = 0\\ 2x - 2 = 0\\ 2x + 1 = 0\end{matrix}\right.$

⇔ $\left[\begin{matrix} x = -2\\ x = 1\\ x = \dfrac{-1}{2}\end{matrix}\right.$

Vậy ` x = -2 ` ; ` x = 1 ` và ` x = \frac{-1}{2} `