x(5 - 2x) + (x - 1) = 15
=> 5x - 2x2 + x - 1 =15
=> 2x2 - 5x - x + 1 = -16
=> 2x2 - 6x + 1 = -16
=> \(2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}+1=-16\)
=> \(2\left(x-\dfrac{3}{2}\right)^2-\dfrac{7}{2}=-16\)
=> \(2\left(x-\dfrac{3}{2}\right)^2=-12,5\)
Vậy x \(\in\)\(\varnothing\)
