`(x+15)/20+(x-7)/4=(x+12)/2+x/5`
`<=>x+15+5(x-7)=10(x+12)+4x`
`<=>x+15+5x-35=10x+120+4x`
`<=>6x-20=14x+120`
`<=>8x=-140`
`<=>-35/2`
Vậy `S={-35/2}`
\(\dfrac{x+15}{20}+\dfrac{x-7}{4}=\dfrac{x+12}{2}+\dfrac{x}{5}\)
\(\Rightarrow\dfrac{x+15}{20}+\dfrac{5\left(x-7\right)}{20}-\dfrac{10\left(x+12\right)}{20}-\dfrac{4x}{20}=0\)
=> x + 15 + 5x - 35 - 10x - 120 - 4x = 0
=> -8x - 140 = 0
=> -8x = 140
=> x = \(-\dfrac{35}{2}\)
Ta có: \(\dfrac{x+15}{20}+\dfrac{x-7}{4}=\dfrac{x+12}{2}+\dfrac{x}{5}\)
\(\Leftrightarrow\dfrac{x+15}{20}+\dfrac{5\left(x-7\right)}{20}=\dfrac{10\left(x+12\right)}{20}+\dfrac{4x}{20}\)
\(\Leftrightarrow x+15+5x-35=10x+120+4x\)
\(\Leftrightarrow6x-20=14x+120\)
\(\Leftrightarrow6x-14x=120+20\)
\(\Leftrightarrow-8x=140\)
hay \(x=-\dfrac{35}{2}\)
Vậy: \(x=-\dfrac{35}{2}\)