5. Tìm x, biết: a) 2x(x – 5) – x(3 + 2x) = 26 b) 3x(x + 2) – 2(x + 2) = 0

`\text{a)}`

`2x(x-5) – x(3 +2x) = 26`

`-> x(2x – 10) – x(3+2x) = 26`

`-> x[ (2x-10) – (3+2x) ] = 26`

`-> x [ (2x – 2x) + (-10-3) ] = 26`

`-> x . (-13) = 26`

`-> x = 26/{-13}`

`-> x = -2` 

Vậy `x = -2`  

`\text{b)}`

`3x(x+2) – 2(x+2) = 0`

`-> (x+2)(3x-2) =0`

`->` \(\left[ \begin{array}{l}x+2 =0\\3x -2 =0\end{array} \right.\)

`->` \(\left[ \begin{array}{l}x=-2\\x = \dfrac{2}{3} \end{array} \right.\)

Vậy `x \in {-2 ; 2/3}`