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\(A=\frac{6}{3.5}+\frac{9}{5.8}+\frac{12}{8.12}+\frac{15}{12.17}\)
\(A=3.\left(\frac{2}{3.5}+\frac{3}{5.8}+\frac{4}{8.12}+\frac{5}{12.17}\right)\)
\(A=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}\right)\)
\(A=3.\left(\frac{1}{3}-\frac{1}{17}\right)< 3.\frac{1}{3}=1\)
=> A < 1
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đề sai thì phải
\(A=\frac{10}{2\cdot12}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{1}{2\cdot3}+\frac{5}{12\cdot17}+\frac{6}{17\cdot23}+\frac{7}{23\cdot30}\)
\(A=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{2}-\frac{1}{3}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(A=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}\)
\(A=\frac{101}{120}\)
\(A=\frac{1}{2.12}+\frac{2}{3.5}+\frac{3}{5.8}+...+\frac{7}{23.30}\)
\(=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}=1-\frac{19}{120}=\frac{101}{120}\)
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\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-...+\dfrac{1}{47}-\dfrac{1}{57}\right)\)
\(=1-\dfrac{18}{57}=\dfrac{39}{57}=\dfrac{13}{19}\)
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\(=\left(\frac{12}{19}.\frac{19}{12}\right).\left(-\frac{13}{17}.\frac{17}{13}\right).\frac{7}{15}\)
\(=1.-1.\frac{7}{15}\)
\(=-1.\frac{7}{15}\)
\(=-\frac{7}{15}\)
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mk ko viết lại đề
=1-3/5+3/5-3/8+...+3/17-3/20
=1-3/20
=17/20
(mk gải thích kĩ hơn chỗ này vì:-3/5+3/5=0 và các số tiếp theo cũng vậy nên cuối cùng chỉ còn 1 và -3/20)
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=\(\left(\frac{5-3}{3.5}\right)+\left(\frac{8-5}{5.8}\right)+\left(\frac{19-8}{8.19}\right)+\left(\frac{32-19}{19.32}\right)+\left(\frac{57-32}{32.57}\right)+\left(\frac{87-57}{57.87}\right)\)
=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{32}+\frac{1}{32}-\frac{1}{57}+\frac{1}{57}-\frac{1}{87}\)
=\(\frac{1}{3}-\frac{1}{87}\)
=\(\frac{28}{87}\)
Ai thấy đúng thì k cho mk nhé!
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= \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{5}{2}.\dfrac{100}{101}\)
= \(\dfrac{250}{101}\)
A = 3 (1/3 - 1/5 + 1/5 - 1/8 + 1/8 - 1/12 + 1/12 - 1/17) = 3(1/3 - 1/17) = 14/17
A = \(\frac{6}{3}.5+\frac{9}{5}.8+\frac{12}{8}.12+\frac{15}{12}.17\)
\(=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{17}\right)\)
\(=3\times\frac{14}{51}\)
\(=\frac{14}{17}\)
CHÚC BẠN HỌC TỐT !!!