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\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-\left[10.\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)\right]=\frac{3}{11}\)
\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(x-\left[10.\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(x-\left[\frac{10}{11}-\frac{10}{55}\right]=\frac{3}{11}\)
\(x-\left[\frac{10}{11}-\frac{2}{11}\right]=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
\(x=\frac{3}{11}+\frac{8}{11}=1\)
\(x-\dfrac{20}{11.13}-\dfrac{20}{23.15}-....-\dfrac{20}{53.55}=\dfrac{3}{11}\)
\(x-\left(\dfrac{20}{11.13}+\dfrac{20}{13.15}+....+\dfrac{20}{53.55}\right)=\dfrac{3}{11}\)
\(x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(x-10\left(\dfrac{1}{11}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(x-\dfrac{8}{11}=\dfrac{3}{11}\)
=> \(x=\dfrac{3}{11}+\dfrac{8}{11}=1\)
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-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
a) \(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
x = 1
b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\) ( nhân cho cả tử và mẫu của các số hạng trên ( ngoại trừ 2/x.(x+1) ) là 2)
\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
=> x + 1 = 18
x = 17
\(a,x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Rightarrow x-\frac{8}{11}=\frac{3}{11}\)
\(\Rightarrow x=1\)
\(b,\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{18}=\frac{1}{x+1}\)
\(\Rightarrow x+1=18\Leftrightarrow x=17\)
Í mà khoan, câu b) mình pít làm đấy, cần mình làm ko pạn??
trong violympic .