`\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}`
