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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
\(S1=\dfrac{5}{10.11}+\dfrac{5}{11.12}+.............+\dfrac{5}{14.15}\)
\(\Leftrightarrow S1=5\left(\dfrac{1}{10.11}+\dfrac{1}{11.12}+...............+\dfrac{1}{14.15}\right)\)
\(\Leftrightarrow S1=5\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+.............+\dfrac{1}{14}-\dfrac{1}{15}\right)\)
\(\Leftrightarrow S1=5\left(\dfrac{1}{10}-\dfrac{1}{15}\right)\)
\(\Leftrightarrow S1=5.\dfrac{1}{30}=\dfrac{1}{6}\)
\(S2=\dfrac{1}{5.9}+\dfrac{1}{9.13}+\dfrac{1}{13.17}+........+\dfrac{1}{21.25}\)
\(\Leftrightarrow4S_2=\dfrac{4}{5.9}+\dfrac{4}{9.13}+..............+\dfrac{4}{21.25}\)
\(\Leftrightarrow4S_2=\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+............+\dfrac{1}{21}-\dfrac{1}{25}\)
\(\Leftrightarrow4S_2=\dfrac{1}{5}-\dfrac{1}{25}\)
\(\Leftrightarrow4S_2=\dfrac{4}{25}\)
\(\Leftrightarrow S_2=\dfrac{16}{25}\)
Đặt
\(A=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+\frac{5}{22.27}+\frac{5}{27.32}+\frac{5}{32.37}\)
\(A=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+\frac{1}{22}-\frac{1}{27}+\frac{1}{27}-\frac{1}{32}+\frac{1}{32}-\frac{1}{37}\)
\(A=\frac{1}{2}-\frac{1}{37}=\frac{35}{74}\)
\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{32}-\frac{1}{37}\)
\(=\frac{1}{2}-\frac{1}{37}\)
\(=\frac{37}{74}-\frac{2}{74}=\frac{35}{74}\)
đề sai thì phải
\(A=\frac{10}{2\cdot12}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{1}{2\cdot3}+\frac{5}{12\cdot17}+\frac{6}{17\cdot23}+\frac{7}{23\cdot30}\)
\(A=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{2}-\frac{1}{3}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(A=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}\)
\(A=\frac{101}{120}\)
\(A=\frac{1}{2.12}+\frac{2}{3.5}+\frac{3}{5.8}+...+\frac{7}{23.30}\)
\(=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}=1-\frac{19}{120}=\frac{101}{120}\)
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 !!!
Sửa:
\(B=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+\frac{5}{22.27}\)
Trả lời
\(B=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}+\frac{5}{22\cdot27}\)
\(\Rightarrow B=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+\frac{1}{22}-\frac{1}{27}\)
\(\Rightarrow B=\frac{1}{2}-\frac{1}{27}\)
\(\Rightarrow B=\frac{25}{54}\)
Vậy B=\(\frac{25}{54}\)
1/11.12 + 1/12.13 + 1/13.14 +...+ 1/43.44
=1/11-1/12+1/12-1/13+1/13-1/14+...+1/43-1/44
=1/11-1/44
=4/44-1/44
=3/44
Ta có A=1/10.11+1/11.12+...+1/98.99+1/99.100
=1/10-1/11+1/11-1/12+...+1/98-1/99+1/99-1/100
=1/10-1/100
=10/100-1/100
=9/100
Vậy A=9/100
Giải:
A=1/10.11+1/11.12+...+1/98.99+1/99.100
A=1/10-1/11+1/11-1/12+...+1/98-1/99+1/99-1/100
A=1/10-1/100
A=9/100
Chúc bạn học tốt!