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Ta có:
\(\dfrac{1}{3}\times\dfrac{12}{12}=\dfrac{12}{36};\)
\(\dfrac{1}{6}\times\dfrac{6}{6}=\dfrac{6}{36};\)
\(\dfrac{1}{10}\times\dfrac{3}{3}=\dfrac{3}{30};\)
\(\dfrac{1}{15}\times\dfrac{2}{2}=\dfrac{2}{30};\)
\(\dfrac{1}{21}\times\dfrac{4}{4}=\dfrac{4}{84};\)
\(\dfrac{1}{28}\times\dfrac{3}{3}=\dfrac{3}{84};\)
\(A=\dfrac{12}{36}+\dfrac{6}{36}+\dfrac{3}{30}+\dfrac{2}{30}+\dfrac{4}{84}+\dfrac{3}{84}+\dfrac{1}{36}\)
\(=\left(\dfrac{12}{36}+\dfrac{6}{36}+\dfrac{1}{36}\right)+\left(\dfrac{3}{30}+\dfrac{2}{30}\right)+\left(\dfrac{4}{84}+\dfrac{3}{84}\right)\)
\(=\dfrac{19}{36}+\dfrac{5}{30}+\dfrac{7}{84}\)
\(=\dfrac{19}{36}+\dfrac{1}{6}+\dfrac{1}{12}\)
\(=\dfrac{19}{36}+\dfrac{6}{36}+\dfrac{3}{36}\)
\(=\dfrac{28}{36}=\dfrac{7}{9}\)
Vậy: \(A=\dfrac{7}{9}\)
=\(\dfrac{1}{3.2}+\dfrac{1}{2.5}+\dfrac{1}{5.3}+\dfrac{1}{3.7}+\dfrac{1}{7.4}+\dfrac{1}{4.9}+\dfrac{1}{9.5}\)=\(\dfrac{1}{3}+\dfrac{1}{5}\)
=\(\dfrac{8}{15}\)
Gọi A = \(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\)
\(\dfrac{1}{2}\)A = \(\dfrac{1}{2}.\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\right)\)
\(\dfrac{1}{2}\)A = \(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(\dfrac{1}{2}\)A = \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(\dfrac{1}{2}\)A = \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\dfrac{1}{2}\)A = \(\dfrac{1}{3}-\dfrac{1}{10}\)
\(\dfrac{1}{2}\)A = \(\dfrac{7}{30}\)
A = \(\dfrac{7}{30}:\dfrac{1}{2}\)
A = \(\dfrac{7}{15}\)
\(\dfrac{x}{6}+\dfrac{x}{10}+\dfrac{x}{15}+........+\dfrac{x}{78}=\dfrac{220}{39}\)
\(\Leftrightarrow\dfrac{2x}{12}+\dfrac{2x}{20}+........+\dfrac{2x}{156}=\dfrac{220}{39}\)
\(\Leftrightarrow2x\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+..........+\dfrac{1}{12.13}\right)=\dfrac{220}{39}\)
\(\Leftrightarrow2x\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{220}{39}\)
\(\Leftrightarrow2x.\dfrac{10}{39}=\dfrac{220}{39}\)
\(\Leftrightarrow x.\dfrac{20}{39}=\dfrac{220}{39}\)
\(\Leftrightarrow x=11\)
Vậy ...
Gọi biểu thức là A
\(A=\dfrac{2x}{12}+\dfrac{2x}{20}+\dfrac{2x}{30}+....+\dfrac{2x}{156}=\dfrac{200}{39}\)
Ta có công thức :
\(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)
Áp dụng công thức trên, ta có :
\(A=\dfrac{2x}{3.4}+\dfrac{2x}{4.5}+\dfrac{2x}{5.6}+....+\dfrac{2x}{12.13}\)
\(A=2x.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{12}-\dfrac{1}{13}\right)\)
\(A=2x.\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)
\(A=2x.\left(\dfrac{10}{39}\right)=\dfrac{200}{39}\)
\(A=2x=\dfrac{200}{39}:\dfrac{10}{39}\)
\(2x=20\)
\(\Rightarrow x=10\)
mink nghĩ vậy bạn ạ
\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{55}+\dfrac{1}{66}\)
\(A=2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\right)\)
\(A=2\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\right)\)
\(A=2\left(\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}-\dfrac{1}{6}\right)+\left(\dfrac{1}{6}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{8}\right)+\left(\dfrac{1}{8}-\dfrac{1}{9}\right)+\left(\dfrac{1}{9}-\dfrac{1}{10}\right)+\left(\dfrac{1}{10}-\dfrac{1}{11}\right)+\left(\dfrac{1}{11}-\dfrac{1}{12}\right)\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{12}\right)\Rightarrow A=\dfrac{1}{3}\)
\(\dfrac{1}{2}N=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(\dfrac{1}{2}N=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\dfrac{1}{2}N=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\)
N=\(\dfrac{2}{5}:\dfrac{1}{2}=\dfrac{4}{5}\)
A =\(2.\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+......+\dfrac{1}{156}\right)\)
A =\(2.\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+..........+\dfrac{1}{12.13}\right)\)
A =2.\(\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)
A=\(2.\dfrac{10}{39}=\dfrac{20}{39}\)
tớ làm hơi gọn nên có gì kho hiểu thì nói tớ