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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) \(\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+....+\frac{1}{8554}\right).x=\frac{31}{94}\)
=> \(\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{91.94}\right).x=\frac{31}{94}\)
=> \(\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}\right)=\frac{31}{94}\)
=> \(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{91}-\frac{1}{94}\right).x=\frac{31}{94}\)
=> \(\frac{1}{3}.\left(1-\frac{1}{94}\right).x=\frac{31}{94}\)
=> \(\frac{1}{3}.\frac{93}{94}.x=\frac{31}{94}\)
=> \(\frac{31}{94}.x=\frac{31}{94}\)
=> \(x=\frac{31}{94}:\frac{31}{94}\)
=> \(x=1\)