`a)` `2-3x=5x+10`
`<=>-3x-5x=10-2`
`<=>-8x=8`
`<=>x=-1`
Vậy `S={-1}`
`b)` `frac{x+5}{3}+frac{2x-7}{4}=frac{x-3}{5}`
`<=>frac{20(x+5)}{60}+frac{15(2x-7)}{60}=frac{12(x-3)}{60}`
`=>20(x+5)+15(2x-7)=12(x-3)`
`<=>20x+100+30x-105=12x-36`
`<=>50x-5=12x-36`
`<=>50x-12x=-36+5`
`<=>38x=-31`
`<=>x=-31/38`
Vậy `S={-31/38}`
`c)` `frac{x-1}{x-3}-frac{1}{x+3}=frac{3x+3}{x^2-9}` Điều kiện xác định: `x\ne±3`
`<=>frac{(x-1)(x+3)-1(x-3)}{(x-3)(x+3)}=frac{3x+3}{(x-3)(x+3)}`
`=>(x-1)(x+3)-1(x-3)=3x+3`
`<=>x^2+3x-x-3-x+3=3x+3`
`<=>x^2+x=3x+3`
`<=>x^2+x-3x-3=0`
`<=>x^2-2x-3=0`
`<=>x^2-3x+x-3=0`
`<=>x(x-3)+(x-3)=0`
`<=>(x-3)(x+1)=0`
`<=>` \(\left[ \begin{array}{l}x-3=0\\x+1=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=3(KTMĐK)\\x=-1(TMĐK)\end{array} \right.\)
Vậy `S={-1}`
