\(\dfrac{x}{2.5}+\dfrac{x}{5.8}+\dfrac{x}{8.11}+\dfrac{x}{11.14}=\dfrac{1}{63}\)
\(x.\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}\right)=\dfrac{1}{63}\)
\(x.\left[\dfrac{1}{3}.\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}\right)\right]=\dfrac{1}{63}\)
\(x.\left[\dfrac{1}{3}.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\right)\right]=\dfrac{1}{63}\)
\(x.\left[\dfrac{1}{3}.\left(\dfrac{1}{2}-\dfrac{1}{14}\right)\right]=\dfrac{1}{63}\)
\(x.\left[\dfrac{1}{3}.\left(\dfrac{7}{14}-\dfrac{1}{14}\right)\right]=\dfrac{1}{63}\)
\(x.\left[\dfrac{1}{3}.\dfrac{3}{7}\right]=\dfrac{1}{63}\)
`x . 1/7 = 1/63`
`x=1/63 : 1/7`
`x=1/63 xx 7`
`x=1/9`
Vậy `x=1/9`
