Tính nhanh: A = \(\dfrac{5}{2.2.4}\)+\(\dfrac{5}{2.4.6}\)+\(\dfrac{5}{2.6.8}\)....+\(\dfrac{5}{2.48.50}\)
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a) \(\dfrac{-5}{9}+\dfrac{3}{5}-\dfrac{3}{9}+\dfrac{-2}{5}\)
=\(\left(\dfrac{-5}{9}-\dfrac{3}{9}\right)+\left(\dfrac{3}{5}+\dfrac{-2}{5}\right)\)
=\(\dfrac{-8}{9}+\dfrac{1}{5}=-\dfrac{31}{45}\)
b) \(\dfrac{5}{17}-\dfrac{9}{15}-\dfrac{2}{17}+\dfrac{-2}{5}\)
=\(\left(\dfrac{5}{17}-\dfrac{2}{17}\right)-\left(\dfrac{9}{15}+\dfrac{2}{5}\right)\)
=\(\dfrac{3}{17}-1=\dfrac{-14}{17}\)
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)
Vậy \(A=\dfrac{1}{2}\)
\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)
Vậy \(B=3\)
= \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{5}{2}.\dfrac{100}{101}\)
= \(\dfrac{250}{101}\)
Ta có: \(A=\dfrac{5}{13}+\dfrac{-5}{7}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}\)
\(=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+\dfrac{-5}{7}\)
\(=1-1+\dfrac{-5}{7}\)
\(=\dfrac{-5}{7}\)
1: \(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
=1/2*10/39
=5/39
2: \(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{11}\right)=\dfrac{5}{2}\cdot\dfrac{10}{11}=\dfrac{50}{22}=\dfrac{25}{11}\)
\(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{7}.\dfrac{1}{9}+\dfrac{5}{9}.\dfrac{3}{7}=\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{3}{7}\right)+\dfrac{5}{7}.\dfrac{1}{9}\\ =\dfrac{5}{9}.\dfrac{4}{7}+\dfrac{5}{7}.\dfrac{1}{9}=\dfrac{20}{63}+\dfrac{5}{63}=\dfrac{25}{63}\)
\(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{3}{7}\)
\(=\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{5}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}.\dfrac{9}{7}\)
\(=\dfrac{5}{7}\)
Em nên thêm hai phân số nữa ở sau đi nha
\(A=\dfrac{5}{2\cdot2\cdot4}+\dfrac{5}{2\cdot4\cdot6}+\dfrac{5}{2\cdot6\cdot8}+...+\dfrac{5}{2\cdot48\cdot50}\)
\(A=\dfrac{5}{2}+\dfrac{5}{6}+\dfrac{5}{2\cdot8}+...+\dfrac{5}{2\cdot48\cdot50}\)
\(A=\dfrac{5}{2+6+2\cdot8+...+2\cdot48\cdot50}\)
\(A=\dfrac{5}{2+6+8+...+48+50}\)
\(A=\dfrac{5}{\left(50-2\right)\div2+1}\)
\(A=\dfrac{5}{25}\)
Vậy \(A=\dfrac{1}{5}\)