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\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\\ \Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\\ \Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\\ 2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\\ \Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{2015}:2\\ \Rightarrow-\frac{1}{x+1}=\frac{2013}{4030}-\frac{1}{2}\\ \Rightarrow-\frac{1}{x+1}=-\frac{1}{2015}\\ \Rightarrow x=2015\)
Vậy x=2015
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\\ \Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\\ \Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\\ 2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\\ \Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{2015}:2\\ \Rightarrow-\frac{1}{x+1}=\frac{2013}{4030}\\ \Rightarrow-\frac{1}{x+1}=-\frac{1}{2015}\\ \Rightarrow x+1=2015\\ \Rightarrow x=2014\)
Vậy x=2014
xin lỗi nhé! vừa nãy mình vội quá nên làm nhầm
Đặt A=1/3+1/6+1/10+...+2/x*(x+1)
1/2A=1/3*2+1/6*2+1/10*2+...+2/2*x*(x+1)
1/2A=1/6+1/12+1/20+...+1/x*(x+1)
1/2A=1/2*3+1/3*4+1/4*5+...+1/x*(x+1)
1/2A=1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/(x+1)
1/2A=1/2-1/x+1
A=(1/2-1/x+1):1/2
A=1-2/x+1
Ta có A=1999/2001
Hay 1-2/x+1=1999/2001
2/x+1=1-1999/2001
2/x+1=2/2001
=>x+1=2001
=>x=2000
Cho A = 1/3+1/6+1/10+...+2/x(x+1)
1/2A= 1/3.2+1/6.2+1/10.2+...+2/x(x+1)2
1/2A= 1/6+1/12+1/20+...+1/x(x+1)
1/2A= 1/2.3+1/3.4+1/4.5+...+1/x(x+1)
1/2A= 1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1
1/2A= 1/2-1/x+1
A = (1/2-1/x+1)/1/2
A = 1-2/x+1
Mà A=1999/2001
=> 1-2/x+1= 1999/2001
2/x+1= 1-1999/2001
2/x+1= 2/2001
=>x+1=2001
=>x = 2000
a) Đặt \(A=\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+.....+\frac{1}{\left(x-2\right)x}+\frac{1}{x\left(x+2\right)}\)
=> \(3A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+.....+\frac{3}{\left(x-2\right)x}+\frac{3}{x\left(x+2\right)}\)
=> \(3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{\left(x-2\right)}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+2}\)
=> 3A = \(\frac{1}{5}-\frac{1}{x+2}\)
=> A = \(\frac{1}{15}-\frac{1}{3x+6}\)
Mà : A = \(\frac{101}{1540}\)
=> \(\frac{1}{15}-\frac{1}{3x+6}=\frac{101}{1540}\)
=> \(\frac{1}{3x+6}=\frac{1}{15}-\frac{101}{1540}=\frac{1}{924}\)
=> 3x + 6 = 924
=> 3(x + 2) = 924
=> x + 2 = 308
=> x = 306
a) Ta có: \({{1} \over x(x+2)}= {{1} \over 3}({{1} \over x}-{{1} \over x+2})\) \(\Rightarrow\) \({{1} \over 3}({{1} \over 5}-{{1} \over 8}+{{1} \over 8}-...+{{1} \over x}-{{1} \over x+2})={{101} \over 1540} \)\(\Leftrightarrow\) \({{1} \over 3}({{1} \over 5}-{{1} \over x+2})={{101} \over 1540}\)\(\Leftrightarrow\)x+2 = 308 \(\Leftrightarrow\) x=306 Lúc sau lm hơi tắt mọi người thông cảm
\(\frac{3}{2}x-\frac{2}{3}=\frac{2}{3}:\frac{3}{2}\)
\(\frac{3}{2}x-\frac{2}{3}=\frac{4}{9}\)
\(\frac{3}{2}x=\frac{4}{9}+\frac{2}{3}\)
\(\frac{3}{2}x=\frac{10}{9}\)
\(x=\frac{10}{9}:\frac{3}{2}\)
\(x=\frac{20}{27}\)
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 ^^
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{999}{1000}\)
\(=\frac{1}{1000}\)
chúc
bn
hk
tốt
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\left[x\left(x+1\right)\right]:2}=\frac{1999}{2001}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)
\(2.\frac{1}{2}-2.\frac{1}{x+1}=\frac{1999}{2001}\)
\(1-\frac{2}{x+1}=\frac{1999}{2001}\)
\(\frac{2}{x+1}=1-\frac{1999}{2001}\)
\(\frac{2}{x+1}=\frac{2}{2001}\)
=> x + 1 = 2001
=> x = 2001 - 1
=> x = 2000
Vậy x = 2000
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\left[x\left(x+1\right)\right]:2}=\frac{1999}{2001}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)
\(2.\frac{1}{2}-2.\frac{1}{x+1}=\frac{1999}{2001}\)
\(1-\frac{2}{x+1}=\frac{1999}{2001}\)
\(\frac{2}{x+1}=1-\frac{1999}{2001}\)
\(\frac{2}{x+1}=\frac{2}{2001}\)
=> x + 1 = 2001
=> x = 2001 - 1
=> x = 2000
Vậy x = 2000