\(x-\dfrac{20}{11\cdot13}-\dfrac{20}{13\cdot15}-...-\dfrac{20}{53\cdot55}=\dfrac{3}{11}\)
=>\(x-10\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{53\cdot55}\right)=\dfrac{3}{11}\)
=>\(x-10\cdot\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
=>\(x-10\cdot\left(\dfrac{1}{11}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
=>\(x-10\cdot\dfrac{4}{55}=\dfrac{3}{11}\)
=>\(x-\dfrac{40}{55}=\dfrac{3}{11}\)
=>\(x-\dfrac{8}{11}=\dfrac{3}{11}\)
=>\(x=1\)
Lời giải:
$x-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}=\frac{3}{11}$
$x-10(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{53.55})=\frac{3}{11}$
$x-10(\frac{13-11}{11.13}+\frac{15-13}{13.15}+...+\frac{55-53}{53.55})=\frac{3}{11}$
$x-10(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55})=\frac{3}{11}$
$x-10(\frac{1}{11}-\frac{1}{15})=\frac{3}{11}$
$x-\frac{8}{33}=\frac{3}{11}$
$x=\frac{3}{11}+\frac{8}{33}=\frac{17}{33}$
