`x-[x/2-[x+3]/4]/2=3-[(1-[6-x]/3). 1/2]/2`
`<=>x-[[2x-x-3]/4]/2=3-[[3-6+x]/3 .1/2]/2`
`<=>x-[2(x-3)]/4=3-[2(x-3)]/6`
`<=>[12x-6(x-3)]/12=[36-4(x-3)]/12`
`<=>12x-6x+18=36-4x+12`
`<=>10x=30`
`<=>x=3`
Vậy `S={3}`
đề hack não v men:V
<=>
\(x-\dfrac{\dfrac{x}{2}-\dfrac{x+3}{4}}{2}-3+\dfrac{\left(1-\dfrac{6-x}{3}\right).\dfrac{1}{2}}{2}=0\)
<=>
\(x-3-\left(\dfrac{\dfrac{2x}{4}-\dfrac{x+3}{4}}{2}-\dfrac{\left(\dfrac{3}{3}-\dfrac{6-x}{3}\right).\dfrac{1}{2}}{2}\right)=0\)
<=>
\(x-3-\left(\dfrac{\dfrac{2x-x-3}{4}}{2}-\dfrac{\left(\dfrac{3-6+x}{3}\right).\dfrac{1}{2}}{2}\right)=0\)
<=>
\(x-3-\left(\dfrac{\dfrac{x-3}{4}}{2}-\dfrac{\left(\dfrac{x-3}{3}\right).\dfrac{1}{2}}{2}\right)=0\)
<=>
\(x-3-\left(\dfrac{\dfrac{x-3}{4}}{2}-\dfrac{\dfrac{x-3}{6}}{2}\right)=0\)
<=>
\(x-3-\left(\dfrac{\dfrac{x-3}{4}-\dfrac{x-3}{6}}{2}\right)=0\)
<=>
\(x-3-\left(\dfrac{\dfrac{6\left(x-3\right)}{24}}{2}-\dfrac{\dfrac{4\left(x-3\right)}{24}}{2}\right)=0\)
<=>
\(x-3-\left(\dfrac{\dfrac{6x-18}{24}}{2}-\dfrac{\dfrac{4x-12}{24}}{2}\right)=0\)
<=>
\(x-3-\left(\dfrac{\dfrac{6x-18-4x+12}{24}}{2}\right)=0\)
<=>
\(x-3-\left(\dfrac{2x-6}{\dfrac{24}{2}}\right)=0\)
<=>
\(x-3-\left(\dfrac{2\left(x-3\right)}{\dfrac{24}{2}}\right)=0\)
<=>
\(x-3-\left(\dfrac{\dfrac{x-3}{12}}{2}\right)=0\)
<=>
\(\dfrac{2x}{2}-\dfrac{6}{2}-\dfrac{x-3}{\dfrac{12}{2}}=0\)
<=>
\(2x-6-\dfrac{x-3}{12}=0\)
<=>
\(\dfrac{24x}{12}-\dfrac{72}{12}-\dfrac{x-3}{12}=0\)
<=>
\(24x-72-x+3=0\)
<=>
\(23x-69=0\)
=> \(x=\dfrac{69}{23}=3\)
