a) Đặt u = \(\dfrac{16x+3}{7}\), ta có:

\(\dfrac{6\left(16x+3\right)}{7}\) - 8 = \(\dfrac{3\left(16x+3\right)}{7}\) + 7

<=> 6.u - 8 = 3.u + 7

=> 6.u - 3.u = 8 + 7

=> 3.u = 15

=> u = 15 / 3

=> u = 5

<=> \(\dfrac{16x+3}{7}\) = 5

=> 16x + 3 = 5 . 7

=> 16x = 35 - 3

=> 16x = 32

=> x = 32 / 16

=> x = 2

Vậy S = { 2 }.