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a) \(\left(31-1\right)\left(\dfrac{-1}{2}x+5\right)=0\)
\(30\left(\dfrac{-1}{2}x+5\right)=0\)
\(\left(\dfrac{-1}{2}x+5\right)=0:30\)
\(\dfrac{-1}{2}x+5=0\)
\(\dfrac{-1}{2}x=0-5\)
\(\dfrac{-1}{2}x=-5\)
\(x=-5:\dfrac{-1}{2}\)
\(x=10\)
b) \(x-\dfrac{20}{11.13}-\dfrac{20}{13.15}-\dfrac{20}{15.17}-...-\dfrac{20}{53.55}=\dfrac{3}{11}\)
\(x-10\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+\dfrac{2}{15.17}+...+\dfrac{2}{53.55}\right)=\dfrac{3}{11}\)
\(x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{17}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\) \(x-10\left(\dfrac{1}{11}-\dfrac{1}{55}\right)=\dfrac{3}{11}\) \(x-10.\dfrac{4}{55}=\dfrac{3}{11}\) \(x-\dfrac{8}{11}=\dfrac{3}{11}\) \(x=\dfrac{3}{11}+\dfrac{8}{11}=1\)
=> 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.( \(\frac{1}{11}\) - \(\frac{1}{13}\) + \(\frac{1}{13}\) - \(\frac{1}{15}\) +...+ \(\frac{1}{53}\) - \(\frac{1}{55}\)) = \(\frac{3}{11}\)
=> x - 10. (\(\frac{1}{11}\) - \(\frac{1}{55}\)) = \(\frac{3}{11}\)
=> x - 10.\(\frac{4}{55}\) = \(\frac{3}{11}\)
=> x - \(\frac{8}{11}\) = \(\frac{3}{11}\)=> x=1 Vậy x=1
\(x-\dfrac{20}{11.13}-\dfrac{20}{23.15}-....-\dfrac{20}{53.55}=\dfrac{3}{11}\)
\(x-\left(\dfrac{20}{11.13}+\dfrac{20}{13.15}+....+\dfrac{20}{53.55}\right)=\dfrac{3}{11}\)
\(x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(x-10\left(\dfrac{1}{11}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(x-\dfrac{8}{11}=\dfrac{3}{11}\)
=> \(x=\dfrac{3}{11}+\dfrac{8}{11}=1\)
Ta có:
x-\(\frac{20}{11.13}\)-\(\frac{20}{13.15}\)-\(\frac{20}{15.17}\)-...-\(\frac{20}{53.57}\)=\(\frac{3}{11}\)
\(\Rightarrow\)\(\frac{20}{11.13}\)-\(\frac{20}{13.15}\)-\(\frac{20}{15.17}\)-...-\(\frac{20}{53.57}\)= x-\(\frac{3}{11}\)
\(\frac{1}{10}\).\((\frac{20}{11.13}.\frac{20}{13.15}.\frac{20}{15.17}...\frac{20}{53.57})\)= x-\(\frac{3}{11}\)
\(\frac{2}{11.13}\).\(\frac{2}{13.15}\).\(\frac{2}{15.17}\)...\(\frac{2}{53.57}\)= x-\(\frac{3}{11}\)
\(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}\)-\(\frac{1}{17}\)+...+\(\frac{1}{53}\)-\(\frac{1}{57}\)=x-\(\frac{3}{11}\)
\(\frac{1}{11}-\frac{1}{57}\)=x-\(\frac{3}{11}\)
\(\frac{46}{627}\)=x-\(\frac{3}{11}\)
x=\(\frac{46}{627}\)-\(\frac{3}{11}\)
Vậy x=\(\frac{-125}{627}\)
\(x-\frac{20}{11.13}-\frac{20}{23.15}-....-\frac{20}{53.55}=\frac{3}{11}\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+....+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
=> \(x=\frac{3}{11}+\frac{8}{11}=1\)