$c$) $6|x-2| = |-15+3|$
$⇔ 6|x-2| = 12$
$⇔ |x-2| =2$
$⇒$ \(\left[ \begin{array}{l}x-2=2\\x-2=-2\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=4\\x=0\end{array} \right.\)
Vậy $x$ $∈$ `{0;4}`
$d$) $x - \dfrac{20}{11.13} - \dfrac{20}{13.15} -\dfrac{20}{15.17} - ... - \dfrac{20}{53.55} = \dfrac{3}{11}$
`⇔ x - (\frac{20}{11.13} + \frac{20}{13.15} +\frac{20}{15.17}+ ... + \frac{20}{53.55}) = \frac{3}{11}`
`⇔ x - 10.(\frac{2}{11.13} + \frac{2}{13.15} +\frac{2}{15.17}+ ... + \frac{2}{53.55}) = \frac{3}{11}`
`⇔ x-10(\frac{1}{11} - \frac{1}{13} + \frac{13} - \frac{1}{15} + \frac{1}{15} - \frac{1}{17} + ... + \frac{1}{53} - \frac{1}{55}) = \frac{3}{11}`
`⇔ x - 10.(\frac{1}{11}- \frac{1}{55}) = \frac{3}{11}`
`⇔ x - \frac{10}{11} + \frac{10}{55} = \frac{3}{11}`
`⇔ x + \frac{10}{55} = \frac{13}{11}`
`⇔ x = \frac{13}{11} - \frac{10}{55}`
`⇔ x = 1`
Vậy `x=1`
