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1 tháng 4 2017
\(\dfrac{\Leftrightarrow3A}{17}=\left(\dfrac{6}{7.13}+\dfrac{9}{13.22}+\dfrac{15}{22.37}+\dfrac{12}{37.49}+\dfrac{30}{49.79}+\dfrac{47}{79.126}\right)x=\dfrac{289}{9}\)
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30 tháng 11 2016
\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)
\(A=17.\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)
\(A=\frac{17}{3}.\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(A=\frac{17}{3}.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)
\(A=\frac{17}{3}.\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(A=\frac{17}{3}.\frac{6}{49}\)
\(B=\frac{39}{7.16}+\frac{65}{16.31}+\frac{52}{31.43}+\frac{26}{43.49}\)
\(B=13.\left(\frac{3}{7.16}+\frac{5}{16.31}+\frac{4}{31.43}+\frac{2}{43.49}\right)\)
\(B=\frac{13}{3}.\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{4}{43.49}\right)\)
\(B=\frac{13}{3}.\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)
\(B=\frac{13}{3}.\left(\frac{1}{7}-\frac{1}{49}\right)=\frac{13}{3}.\frac{6}{49}\)
\(\frac{A}{B}=\frac{\frac{17}{3}.\frac{6}{49}}{\frac{13}{3}.\frac{6}{49}}=\frac{17}{13}\)
SK
8 tháng 1 2018
\(A=17\left(\frac{2}{7\cdot13}+\frac{3}{13\cdot22}+\frac{5}{22\cdot37}+\frac{4}{37\cdot49}\right)\)
\(=\frac{17}{3}\left(\frac{6}{7\cdot13}+\frac{9}{13\cdot22}+\frac{15}{22\cdot37}+\frac{12}{37\cdot49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(B=13\left(\frac{3}{7\cdot16}+\frac{5}{16\cdot31}+\frac{4}{31\cdot43}+\frac{2}{43\cdot49}\right)\)
\(=\frac{13}{3}\left(\frac{9}{7\cdot16}+\frac{15}{16\cdot31}+\frac{12}{31\cdot43}+\frac{6}{43\cdot49}\right)\)
\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)
\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}{\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}\)\(=\frac{\frac{17}{3}}{\frac{13}{3}}=\frac{17}{13}\)
\(\dfrac{3}{17}A=\dfrac{6}{7.13}+\dfrac{9}{13.22}+\dfrac{15}{22.37}+\dfrac{12}{37.49}\)
\(\dfrac{3}{17}A=\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{49}\)
\(\dfrac{3}{17}A=\dfrac{1}{7}-\dfrac{1}{49}=\dfrac{6}{49}\)
\(A=\dfrac{6}{49}:\dfrac{3}{17}=\dfrac{6.17}{49.3}=\dfrac{34}{49}\)