Question

[Lê Duy Bảo]

Cho A= \(\frac{1}{5.8}\)+ \(\frac{1}{8.11}\)+ . . . + \(\frac{1}{29.32}\)+ \(\frac{1}{32.35}\)

       B= \(\frac{1}{5.10}\)+ \(\frac{1}{10.15}\)+. . . + \(\frac{1}{25.30}\)+ \(\frac{1}{30.35}\)

Tìm x biết:

                     \(\frac{\left(1\frac{1}{5}-\frac{4}{9}\right).\left(3^{22}+4.3^{18}\right)}{9^8.\left(x-2\right)}\)  = \(\frac{A}{B}\)

Answer 1

[Sherlockichi Kudoyle]

\(A=\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{29.32}+\frac{1}{32.35}\)

\(3A=\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{32.35}\)

\(3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{32}-\frac{1}{35}\)

\(3A=\frac{1}{5}-\frac{1}{35}=\frac{6}{35}\)

\(A=\frac{6}{35}.\frac{1}{3}=\frac{2}{35}\)

\(B=\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{30.35}\)

\(5B=\frac{5}{5.10}+\frac{5}{10.15}+....+\frac{5}{30.35}\)

\(5B=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{30}-\frac{1}{35}\)

\(5B=\frac{1}{5}-\frac{1}{35}=\frac{6}{35}\)

\(B=\frac{6}{35}.\frac{1}{5}=\frac{6}{175}\)

\(\frac{\left(\frac{6}{5}-\frac{4}{9}\right).\left(3^{22}+4.3^{18}\right)}{9^8.\left(x-2\right)}=\frac{\frac{2}{35}}{\frac{6}{175}}\)

\(\frac{\frac{34}{45}.3^{18}.\left(3^4+4\right)}{\left(3^2\right)^8.\left(x-2\right)}=\frac{5}{3}\)

\(\frac{\frac{34}{45}.85.3^{18}}{3^{16}.\left(x-2\right)}=\frac{5}{3}\)

\(\frac{\frac{578}{9}.3^2}{x-2}=\frac{5}{3}\)

\(\frac{578}{x-2}=\frac{5}{3}\)

\(\Rightarrow578.3=5x-10\)

\(\Rightarrow1734+10=5x\)

\(\Rightarrow x=\frac{1744}{5}=348,8\)

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