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\(\frac{1}{6}\) + \(\frac{2}{15}\) + \(\frac{3}{40}\) + \(\frac{4}{96}\) + \(\frac{5}{204}\)
= \(\frac{3}{10}\)+\(\frac{3}{40}\)+ \(\frac{4}{96}\) + \(\frac{5}{204}\)
=\(\frac{3}{8}\)+\(\frac{4}{96}\)+ \(\frac{5}{204}\)
= \(\frac{5}{12}\)+ \(\frac{5}{204}\)
Hok tốt !
k nha <3
= \(\frac{15}{34}\)
\(A=\frac{3-2}{2\times3}+\frac{5-3}{3\times5}+\frac{8-5}{5\times8}+...\frac{38-30}{30\times38}+\frac{47-38}{38\times47}\)
\(A=\frac{3}{2\times3}-\frac{2}{2\times3}+\frac{5}{3\times5}-\frac{3}{3\times5}+...\frac{38}{30\times38}-\frac{30}{30\times38}+\frac{47}{38\times47}-\frac{38}{38\times47}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}+...+\frac{1}{30}-\frac{1}{38}+\frac{1}{38}-\frac{1}{47}\)
\(A=\frac{1}{2}-\frac{1}{47}=\frac{47}{94}-\frac{2}{94}=\frac{45}{94}\)
Lời giải:
Tổng 10 phân số đầu tiên là:
$\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}+\frac{5}{204}+.....+\frac{10}{2679}$
$=\frac{1}{2.3}+\frac{2}{3.5}+\frac{3}{5.8}+\frac{5}{8.12}+\frac{5}{12.17}+\frac{6}{17.23}+\frac{7}{23.30}+\frac{8}{30.38}+\frac{9}{38.47}+\frac{10}{47.57}$
$=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{47}-\frac{1}{57}$
$=\frac{1}{2}-\frac{1}{57}=\frac{55}{114}$
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a) Ta có B = \(\left(\frac{2}{15}+\frac{3}{40}+\frac{4}{96}+\frac{5}{204}+\frac{6}{391}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\left(\frac{2}{3.5}+\frac{3}{5.8}+\frac{4}{8.12}+\frac{5}{12.17}+\frac{6}{17.23}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{23}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\left(\frac{1}{3}-\frac{1}{23}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\frac{20}{69}.x.\left(x-1\right)=\frac{20}{69}\)
=> \(x.\left(x-1\right)=\frac{20}{69}:\frac{20}{69}\)
=> \(x.\left(x-1\right)=1\)
=> \(x\in\varnothing\)
a) \(\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+....+\frac{1}{8554}\right).x=\frac{31}{94}\)
=> \(\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{91.94}\right).x=\frac{31}{94}\)
=> \(\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}\right)=\frac{31}{94}\)
=> \(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{91}-\frac{1}{94}\right).x=\frac{31}{94}\)
=> \(\frac{1}{3}.\left(1-\frac{1}{94}\right).x=\frac{31}{94}\)
=> \(\frac{1}{3}.\frac{93}{94}.x=\frac{31}{94}\)
=> \(\frac{31}{94}.x=\frac{31}{94}\)
=> \(x=\frac{31}{94}:\frac{31}{94}\)
=> \(x=1\)