\(\Leftrightarrow5-x+6=12-8x\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(5-\left(x-6\right)=4\left(3-2x\right)\\ 5-x+6=4\cdot3-4\cdot2x\\ 5-x+6=12-8x\\ \Leftrightarrow-x+8x=12-5-6\\ \Leftrightarrow7x=1\\ \Leftrightarrow x=\dfrac{1}{7}\)
`5-(x-6) = 4( 3-2x)`
`<=> 5-x+6 = 12 - 8x`
`<=> 11 - x = 12 - 8x`
`<=> 11 = 12 - 8x + x`
`<=> 11 = 12 - 7x`
`<=> 7x = 12 - 11`
`<=> 7x=1`
`<=> x = 1/7(tm)`
Vậy `S={1/7}`