`(x - 2)/3 = (x + 1)/4`
`<=> (x - 2) . 3 = (x + 1) . 4`
`<=> 4x . 3 = (x + 1) . 4`
`<=> (4 - 3)x = 11`
`<=> x = 11`
Vậy `x = 11`
`[ x - 2 ] / 3 = [ x + 1 ] / 4`
`4 ( x - 2 ) = 3 ( x + 1 )`
`4x - 8 = 3x + 3`
`4x - 3x = 3 + 8`
`x = 11`
\(\Leftrightarrow4\left(x-2\right)=3\left(x+1\right)\)
\(\Leftrightarrow4x-8-3x-3=0\)
\(\Leftrightarrow x-11=0\)
\(\Leftrightarrow x=11\)