(3.5)^2 :(8^5.8^6)=?
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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 !!!
Ta có: \(\dfrac{2}{3\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{16}{122\cdot138}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{122}-\dfrac{1}{138}\)
\(=\dfrac{1}{3}-\dfrac{1}{138}\)
\(=\dfrac{46}{138}-\dfrac{1}{138}\)
\(=\dfrac{45}{138}=\dfrac{15}{46}\)
Giải:
2/3.5+3/5.8+...+16/122.138
=1/3-1/5+1/5-1/8+...+1/122-1/138
=1/3-1/138
=45/138=15/46
Chúc bạn học tốt!
\(A=\frac{6}{3.5}+\frac{9}{5.8}+\frac{12}{8.12}+\frac{15}{12.17}\)
\(A=3.\left(\frac{2}{3.5}+\frac{3}{5.8}+\frac{4}{8.12}+\frac{5}{12.17}\right)\)
\(A=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}\right)\)
\(A=3.\left(\frac{1}{3}-\frac{1}{17}\right)< 3.\frac{1}{3}=1\)
=> A < 1
\(\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{19.32}+\frac{25}{32.57}+\frac{30}{57.87}\)
\(=\frac{5-3}{3.5}+\frac{8-5}{3}+\frac{19-8}{8.19}+\frac{32-29}{19.32}+\frac{57-32}{32.57}+\frac{87-57}{57.87}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{32}+\frac{1}{32}-\frac{1}{57}+\frac{1}{57}-\frac{1}{87}\)
\(=\frac{1}{3}-\frac{1}{87}=\frac{28}{87}\)
Ta có :
\(A=\frac{6}{3.5}+\frac{9}{5.8}+\frac{12}{8.12}+\frac{15}{12.17}\)
\(A=3.\left(\frac{2}{3.5}\right)+3.\left(\frac{3}{5.8}\right)+3.\left(\frac{4}{8.12}\right)+3.\left(\frac{5}{12.17}\right)\)
\(A=3.\left(\frac{2}{3.5}+\frac{3}{5.8}+\frac{4}{8.12}+\frac{5}{12.17}\right)\)
\(A=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}\right)\)
\(A=3.\left(\frac{1}{3}-\frac{1}{17}\right)\)
\(A=3.\frac{14}{51}\)
\(A=\frac{14}{17}< 1\)
Vậy A < 1
_Chúc bạn học tốt_
đề sai thì phải
\(A=\frac{10}{2\cdot12}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{1}{2\cdot3}+\frac{5}{12\cdot17}+\frac{6}{17\cdot23}+\frac{7}{23\cdot30}\)
\(A=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{2}-\frac{1}{3}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(A=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}\)
\(A=\frac{101}{120}\)
\(A=\frac{1}{2.12}+\frac{2}{3.5}+\frac{3}{5.8}+...+\frac{7}{23.30}\)
\(=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}=1-\frac{19}{120}=\frac{101}{120}\)