`(1/(1×3)+1/(3×5)+...+1/(23×25))×[(x+1)+(x+3)+(x+5)+...+(x+23)]`
Đặt `A=1/(1×3)+1/(3×5)+...+1/(23×25)`
`=>A×(x+x+...+x)+(1+3+5+...+23)=144` ( có `(23-1):2+1=12` số `x` )
`=>A×12x=144-[(23+1)×12:2]`
`=>A×12x=144-144`
`=>A×12x=0`
Ta có : `A=1/(1×3)+1/(3×5)+1/(23×25)`
`=>2A=(1/(1×3)+1/(3×5)+...+1/(23×25))×2`
`=>2A=2/(1×3)+2/(3×5)+....+2/(23×25)`
`=>2A=(3-1)/(1×3)+(5-3)/(3×5)+...+(25-23)/(23×25)`
`=>2A=3/(1×3)-1/(1×3)+5/(3×5)-3/(3×5)+...+25/(23×25)-23/(23×25)`
`=>2A=1-1/3+1/3-1/5+...+1/23-1/25`
`=>2A=1-1/25`
`=>2A=24/25`
`=>A=12/25`
`=>A×12x=0`
`=>24/25×12x=0`
`=>12x=0:24/25`
`=>12x=0`
`=>x=0`
Vậy `x=0`
