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\(x-\dfrac{20}{11.13}-\dfrac{20}{23.15}-....-\dfrac{20}{53.55}=\dfrac{3}{11}\)
\(x-\left(\dfrac{20}{11.13}+\dfrac{20}{13.15}+....+\dfrac{20}{53.55}\right)=\dfrac{3}{11}\)
\(x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(x-10\left(\dfrac{1}{11}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(x-\dfrac{8}{11}=\dfrac{3}{11}\)
=> \(x=\dfrac{3}{11}+\dfrac{8}{11}=1\)
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=\(\frac{-125}{627}\)
\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-\left[10.\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)\right]=\frac{3}{11}\)
\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(x-\left[10.\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(x-\left[\frac{10}{11}-\frac{10}{55}\right]=\frac{3}{11}\)
\(x-\left[\frac{10}{11}-\frac{2}{11}\right]=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
\(x=\frac{3}{11}+\frac{8}{11}=1\)
=> 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 - 20/11 . 13 - 20/13 . 15 - ...- 20/53 . 55 = 25/11
X - ( 20/11 . 13 + 20/13 . 15 + ...+ 20/53 . 55 ) = 25/11
X - 10 . ( 2/11 . 13 + 2/13 . 15 + ...+ 2/53 . 55 ) = 25/11
X - 10. ( 1/11 - 1/13 + 1/13 - 1/15 + ...+ 1/53 - 1/55 ) = 25/11
X - 10. ( 1/11 - 1/55 ) = 25/11
X - 10 . 4/55 = 25/11
X - 8/11 = 25/11
X = 25/11 + 8/11
X = 33/11
X = 3
Tham khảo nha !!!
\(x-\frac{20}{11\times13}-\frac{20}{13\times15}-...-\frac{20}{53\times5}=\frac{25}{11}\)
\(\Rightarrow x-\frac{\left(\frac{5}{11}+\frac{5}{13}-\frac{5}{13}+\frac{5}{15}-\frac{5}{15}-...-\frac{5}{53}+\frac{5}{55}\right)}{6}=\frac{25}{11}\)
\(\Rightarrow x-\frac{\left(\frac{5}{11}+\frac{5}{55}\right)}{6}=\frac{25}{11}\)
\(\Rightarrow x=\frac{25}{11}+\frac{\left(\frac{5}{11}+\frac{5}{55}\right)}{6}\)
\(\Rightarrow x=\frac{26}{11}\)
x-10( 2/11.13+2/13.15+...+2/53.55)=3/11
x-10(1/11-1/55)=3/11
x-10.4/55=3/11
x-40/55=3/11
x=3/11+40/55
x= 1
x-10.(2/11.13+2/13.15+....+2/53.55)=3/11
x-10.(1/11-1/13+...+1/53-1/55)=3/11
x-10.(1/11-1/55)=3/11
x-10.4/55=3/11
x-8/11=3/11
x=1
Nhớ k cho mình nhé