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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)  \(\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+....+\frac{1}{8554}\right).x=\frac{31}{94}\)

=> \(\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{91.94}\right).x=\frac{31}{94}\) 

=> \(\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}\right)=\frac{31}{94}\)

=> \(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{91}-\frac{1}{94}\right).x=\frac{31}{94}\)

=> \(\frac{1}{3}.\left(1-\frac{1}{94}\right).x=\frac{31}{94}\)

=> \(\frac{1}{3}.\frac{93}{94}.x=\frac{31}{94}\)

=> \(\frac{31}{94}.x=\frac{31}{94}\)

=> \(x=\frac{31}{94}:\frac{31}{94}\)

=> \(x=1\)

=1/2x3+2/3x5+3/5x8+4/8x12+5/12x17+6/17x23+7/23x30

=1/2-1/3+1/3-1/5+...+1/23-1/30

=1/2-1/30

=7/15

\(A=\frac{3-2}{2\times3}+\frac{5-3}{3\times5}+\frac{8-5}{5\times8}+...\frac{38-30}{30\times38}+\frac{47-38}{38\times47}\)

\(A=\frac{3}{2\times3}-\frac{2}{2\times3}+\frac{5}{3\times5}-\frac{3}{3\times5}+...\frac{38}{30\times38}-\frac{30}{30\times38}+\frac{47}{38\times47}-\frac{38}{38\times47}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}+...+\frac{1}{30}-\frac{1}{38}+\frac{1}{38}-\frac{1}{47}\)

\(A=\frac{1}{2}-\frac{1}{47}=\frac{47}{94}-\frac{2}{94}=\frac{45}{94}\)

\(\frac{1}{6}\) + \(\frac{2}{15}\) + \(\frac{3}{40}\) + \(\frac{4}{96}\) + \(\frac{5}{204}\)

= \(\frac{3}{10}\)+\(\frac{3}{40}\)+ \(\frac{4}{96}\) + \(\frac{5}{204}\)

=\(\frac{3}{8}\)+\(\frac{4}{96}\)+ \(\frac{5}{204}\)

= \(\frac{5}{12}\)+ \(\frac{5}{204}\)

Hok tốt !

k nha <3

= \(\frac{15}{34}\)

1)\(\frac{252}{x}=\frac{84}{97}\Rightarrow\)\(\frac{84}{97}=\frac{252}{291}\Rightarrow x=291\)

6) \(\frac{y}{15}=\frac{2}{5}\Rightarrow\frac{2}{5}=\frac{6}{15}\Rightarrow x=6\)

= 625/1224

Lời giải:

Tổng 10 phân số đầu tiên là:
$\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}+\frac{5}{204}+.....+\frac{10}{2679}$

$=\frac{1}{2.3}+\frac{2}{3.5}+\frac{3}{5.8}+\frac{5}{8.12}+\frac{5}{12.17}+\frac{6}{17.23}+\frac{7}{23.30}+\frac{8}{30.38}+\frac{9}{38.47}+\frac{10}{47.57}$

$=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{47}-\frac{1}{57}$

$=\frac{1}{2}-\frac{1}{57}=\frac{55}{114}$