`C_1:`
`(9+x)/(13-x)=5/6`
Vì `9+x` rút gọn `=5` nên `9+x \vdots 5 < 10`
`=>x={1;6}`
Vì `13-x=6` nhưng vì `9+x` rút gọn nên `13-x \vdots 6 <10`
`=>x=1`
Ta có : `9+x∩13-x=1`
Vậy `x=1`
`C_2:`
`(9+x)/(13-x)=5/6`
`=>(9+x)/(13-x)=5/6`
`=>(9+x)6=(13-x)5`
`=>54+6x=65-5x`
`=>6x+5x=65-54`
`=>11x=11`
`=>x=11/11=1`