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\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{x+1-2}{2\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow2003.2\left(x+1\right)=4010\left(x-1\right)\)
\(\Leftrightarrow4006x+4006=4010x-4010\)
\(\Leftrightarrow-4x=-8016\)
\(\Leftrightarrow x=2004\)
Vậy x = 2004
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2003}{2005}\)
\(\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}\right).\frac{1}{2}=\frac{2003}{2005}.\frac{1}{2}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{2}{x.\left(x+1\right).2}=\frac{2003}{4020}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2003}{4020}\)
\(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{\left(x+1\right)-x}{x.\left(x+1\right)}=\frac{2003}{4020}\)
\(\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+...+\frac{x+1}{\left(x+1\right).x}-\frac{x}{\left(x+1\right).x}=\frac{2003}{4020}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{2003}{4020}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4020}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4020}=\frac{7}{4020}\)
\(\frac{7}{\left(x+1\right).7}=\frac{7}{4020}\)
\(\left(x+1\right).7=4020\)
\(\Rightarrow x=....\)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2003}{2005}\)
\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{4008}{2005}\)
\(2.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)
\(2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{4008}{2005}\)
\(=>2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4008}{2005}\)
\(2.\left(1-\frac{1}{x+1}\right)=\frac{4008}{2005}\)
=> \(1-\frac{1}{x+1}=\frac{4008}{2005}:2=\frac{2004}{2005}\)
\(\frac{1}{x+1}=1-\frac{2004}{2005}=\frac{1}{2005}\)
=>x+1=2005
=>x=2004
Ta có :
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(1+1-\frac{2}{x+1}=\frac{2003}{2005}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=2-\frac{2003}{2005}\)
\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2007}{2005}\)
\(\Leftrightarrow\)\(x+1=2:\frac{2007}{2005}\)
\(\Leftrightarrow\)\(x+1=\frac{4010}{2007}\)
\(\Leftrightarrow\)\(x=\frac{4010}{2007}-1\)
\(\Leftrightarrow\)\(x=\frac{2003}{2007}\)
Vậy \(x=\frac{2003}{2007}\)
Chúc bạn học tốt ~
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{2\times3}+\frac{2}{3\times4}+\frac{2}{4\times5}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}+...+\frac{2}{x}-\frac{2}{x+1}=\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{2}-\frac{2}{x+1}=\frac{2005}{2007}\)
\(\Rightarrow1-\frac{2}{x+1}=\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{x+1}=1-\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{2007}\)
\(\Rightarrow x+1=2007\)
\(\Rightarrow x=2006\)
\(\frac{1}{2}\cdot\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}\cdot\frac{2005}{2007}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2005}{4014}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2005}{4014}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2005}{4014}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2005}{4014}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2007}\)
\(\Rightarrow x+1=2007\)
\(x=2007-1\)
\(x=2006\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{4}{9}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{2}{3}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
Vậy x=\(\frac{20}{27}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=1-\frac{4}{5}\)
\(\left(\frac{9}{11}-x\right):\frac{-10}{11}=\frac{1}{5}\)
\(\frac{9}{11}-x=\frac{1}{5}\cdot\frac{-10}{11}\)
\(\frac{9}{11}-x=\frac{-2}{11}\)
\(x=\frac{9}{11}-\frac{-2}{11}\)
\(x=1\)
Vậy x=1
\(\frac{-11}{12}\cdot x+\frac{3}{4}=\frac{-1}{6}\)
\(\frac{-11}{12}\cdot x=\frac{-1}{6}-\frac{3}{4}\)
\(\frac{-11}{12}\cdot x=\frac{21}{12}\)
\(x=\frac{-21}{11}\)
Vậy x=\(\frac{-21}{11}\)
\(\frac{-5}{4}-\left(1\frac{1}{2}+x\right)=4,5\)
\(\frac{3}{2}+x=\frac{-5}{4}-\frac{9}{2}\)
\(\frac{3}{2}+x=\frac{23}{4}\)
\(x=\frac{17}{4}\)
Vậy x=\(\frac{17}{4}\)
\(\left(\frac{3}{4}-x:\frac{2}{15}\right)\cdot\frac{1}{5}=-2,6\)
\(\frac{3}{4}-x:\frac{2}{15}=\frac{-13}{5}:\frac{1}{5}\)
\(\frac{3}{4}-x:\frac{2}{15}=-13\)
\(x:\frac{2}{15}=\frac{3}{4}-\left(-13\right)\)
\(x:\frac{2}{15}=\frac{45}{4}\)
\(x=\frac{3}{2}\)
Vậy x=\(\frac{3}{2}\)
\(3-\left(\frac{1}{6}-x\right)\cdot\frac{2}{3}=\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=\frac{2}{3}:\frac{2}{3}\)
\(3-\left(\frac{1}{6}-x\right)=1\)
\(\frac{1}{6}-x=2\)
\(x=\frac{1}{6}-2\)
\(x=\frac{-11}{6}\)
Vậy x=\(\frac{-11}{6}\)
\(\left(1-2x\right)\cdot\frac{4}{5}=\left(-2\right)^3\)
\(1-2x=\frac{-1}{10}\)
\(2x=1-\frac{-1}{10}\)
\(2x=\frac{11}{10}\)
\(x=\frac{11}{20}\)
Vậy x=\(\frac{11}{20}\)
\(\frac{1}{6}-\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{1}{8}\)
\(\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{7}{12}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{7}{12}\) \(\frac{1}{2}x-\frac{1}{3}=\frac{-7}{12}\)
\(\frac{1}{2}x=\frac{11}{12}\) \(\frac{1}{2}x=\frac{-1}{4}\)
\(x=\frac{11}{6}\) \(x=\frac{-1}{2}\)
Vậy \(x\in\left\{\frac{11}{6};\frac{-1}{2}\right\}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{6}{9}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
tk mình đi mình làm nốt cho hjhj ^^
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4010}=\frac{1}{2005}\)
\(\Rightarrow x+1=2005\Rightarrow x=2004\)