\(\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2}}}}=\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{\dfrac{3}{2}}}}=\dfrac{1}{2-\dfrac{1}{2-\dfrac{2}{3}}}=\dfrac{1}{2-\dfrac{1}{\dfrac{4}{3}}}=\dfrac{1}{2-\dfrac{3}{4}}=\dfrac{1}{\dfrac{5}{4}}=\dfrac{4}{5}\)
\(\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2}}}}=\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{\dfrac{3}{2}}}}=\dfrac{1}{2-\dfrac{1}{2-\dfrac{2}{3}}}=\dfrac{1}{2-\dfrac{1}{\dfrac{4}{3}}}=\dfrac{1}{2-\dfrac{3}{4}}=\dfrac{1}{\dfrac{5}{4}}=\dfrac{4}{5}\)
Thực hiện phép tính :
a ) \(1-\dfrac{1}{2}:\dfrac{-3}{2}+\dfrac{2}{3}\)
b ) \(\dfrac{-3}{4}-\dfrac{-7}{2}+\dfrac{2}{-3}\)
c ) \(0,25-3\dfrac{1}{2}:\dfrac{1}{2}+\dfrac{-3}{4}.\dfrac{2}{-3}\)
Chứng minh rằng:
a) \(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...\dfrac{2018}{2019!}\)<1
b) \(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+...+\dfrac{999.1000-1}{1000!}\)<2
Cho \(A=\left(\dfrac{2}{2^2}-1\right).\left(\dfrac{1}{3^2}-1\right).\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)So sánh A với \(-\dfrac{1}{2}\)
11) \(\dfrac{1}{3}\)x + \(\dfrac{1}{5}\) - \(\dfrac{1}{2}\)x = 1\(\dfrac{1}{4}\)
12) \(\dfrac{3}{2}\)( x + \(\dfrac{1}{2}\)) - \(\dfrac{1}{8}\)x = \(\dfrac{1}{4}\)
13) \(\dfrac{3}{5}\)( x + \(\dfrac{1}{3}\)) - \(\dfrac{1}{3}\)( x - \(\dfrac{1}{2}\)) = 1\(\dfrac{1}{3}\)
14) \(\dfrac{2}{5}\)( x + \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)( x - \(\dfrac{1}{5}\))
Giup mik nha mik cần gấp
Tìm x, biết:
a) \(\dfrac{-3}{7}+x=\dfrac{1}{3}\)
b) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
c) \(1\dfrac{1}{3}:0,8=\dfrac{2}{3}:0,1x\)
d) \(\left|x-3\right|=\dfrac{1}{2}\)
\(\dfrac{0,75+\dfrac{3}{5}-\dfrac{3}{7}-\dfrac{3}{13}}{2,75+2\dfrac{1}{5}-\dfrac{11}{7}-\dfrac{11}{3}}-\dfrac{0,75-\dfrac{3}{11}-\dfrac{1}{3}}{\dfrac{5}{22}-0,625+\dfrac{5}{18}}\)
Tìm x:
a) \(\dfrac{2}{3}\) - \(\dfrac{1}{3}\) . ( x - \(\dfrac{3}{2}\) ) - \(\dfrac{1}{2}\) . ( 2x + 1 ) = 5
b) ( x + \(\dfrac{1}{2}\) ) . ( \(\dfrac{3}{4}\) - x ) = 0
c) \(\dfrac{2x-1}{-3+2}\) = 0
1+\(\dfrac{1}{1+2}\)+\(\dfrac{1}{1+2+3}\)+...+\(\dfrac{1}{1+2+3+...+2012}\)
Tìm x biết:
a) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
b)\(\dfrac{x+4}{2010}+\dfrac{x+3}{2010}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)