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\(A=\frac{20}{1\cdot6}+\frac{20}{6\cdot11}+...+\frac{20}{51\cdot56}+\frac{20}{56\cdot61}\)
\(A=\frac{20}{5}\cdot\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{51}-\frac{1}{56}+\frac{1}{56}-\frac{1}{61}\right)\)
\(A=4\cdot\left(1-\frac{1}{61}\right)\)
\(A=4\cdot\frac{60}{61}\)
\(A=\frac{240}{61}\)
\(A=\frac{20}{1.6}+\frac{20}{6.11}+...+\frac{20}{51.56}+\frac{20}{56.61}\)
\(A=\frac{20}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{51}-\frac{1}{56}+\frac{1}{56}-\frac{1}{61}\right)\)
\(A=4.\left(1-\frac{1}{61}\right)\)
\(A=4.\frac{60}{61}=\frac{240}{61}\)
Bài 1: Ta có:
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)
\(=\dfrac{20}{8.14}+\dfrac{20}{14.20}+\dfrac{20}{20.26}+\dfrac{20}{26.32}\)
\(=\dfrac{20}{6}\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+\dfrac{6}{26.32}\right)\)
\(=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+...+\dfrac{1}{26}-\dfrac{1}{32}\right)\)
\(=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{32}\right)=\dfrac{20}{6}.\dfrac{3}{32}=\dfrac{5}{16}\)
Vậy \(M=\dfrac{5}{16}\)
Bài 2: Ta có:
\(A=\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+...+\dfrac{1}{210}\)
\(=\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+...+\dfrac{1}{14.15}\)
\(=\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{6}-\dfrac{1}{15}=\dfrac{1}{10}\)
Vậy \(A=\dfrac{1}{10}\)
Giải:
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}.\)
\(M=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+\dfrac{5}{208}.\)
\(M=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+\dfrac{5}{13.16}.\)
\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right).\)
\(M=\dfrac{5}{3}\left[\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+\left(\dfrac{1}{10}-\dfrac{1}{10}\right)+\left(\dfrac{1}{13}-\dfrac{1}{13}\right)+\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\right].\)
\(M=\dfrac{5}{3}\left[0+0+0+\left(\dfrac{1}{4}-\dfrac{1}{16}\right).\right]\)
\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{16}\right).\)
\(M=\dfrac{5}{3}\left(\dfrac{4}{16}-\dfrac{1}{16}\right).\)
\(M=\dfrac{5}{3}.\dfrac{3}{16}.\)
\(M=\dfrac{15}{48}=\dfrac{5}{16}.\)
a, sai đề
b, \(\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\Rightarrow\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\) ( nhân cả 2 vế với \(\dfrac{1}{2}\) )
\(\Rightarrow\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\Rightarrow x+1=18\Rightarrow x=17\)
Vậy x = 17
Câu a thiếu đề rồi bạn ơi mik giải câu b đây:
\(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)
\(2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(2\left(\dfrac{1}{6}-\dfrac{1}{x+2}\right)=\dfrac{2}{9}\)
\(\dfrac{1}{6}-\dfrac{1}{x+2}=\dfrac{2}{9}:2\)
\(\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Rightarrow x+1=18\Rightarrow x=17\)
Vậy x = 17
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\\ =\dfrac{20}{8.14}+\dfrac{20}{14.20}+\dfrac{20}{20.26}+\dfrac{20}{26.32}\\ =\dfrac{20}{6}.\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+\dfrac{6}{26.32}\right)\\ =\dfrac{20}{6}.\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{32}\right)\\ =\dfrac{20}{6}.\left(\dfrac{1}{8}-\dfrac{1}{32}\right)\\ =\dfrac{20}{6}.\dfrac{3}{32}=\dfrac{5}{16}\)
Anh giải cách đây 3 ngày rồi!
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\\ M=\dfrac{20}{8.14}+\dfrac{20}{14.20}+\dfrac{20}{20.26}+\dfrac{20}{26.32}\\ M=\dfrac{20}{6}\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+\dfrac{6}{26.32}\right)\\M=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{32}\right)\\ M=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{32}\right)=\dfrac{20}{6}.\dfrac{3}{32}=\dfrac{5}{16}\)
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)
\(\Rightarrow M=\dfrac{20}{8.14}+\dfrac{20}{14.20}+\dfrac{20}{20.26}+\dfrac{20}{26.32}\)
\(\Rightarrow M=\dfrac{20}{6}\left(\dfrac{6}{8.14}+\dfrac{8}{14.20}+\dfrac{8}{20.26}+\dfrac{8}{26.32}\right)\)
\(\Rightarrow M=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{32}\right)\)
\(\Rightarrow M=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{32}\right)\)
\(\Rightarrow M=\dfrac{20}{6}.\dfrac{3}{32}=\dfrac{5}{16}\)
\(A=\dfrac{20}{1.6}+\dfrac{20}{6.11}+.........+\dfrac{20}{51.56}+\dfrac{20}{56.61}\)
\(\dfrac{1}{4}A=\dfrac{5}{1.6}+\dfrac{5}{6.11}+........+\dfrac{5}{51.56}+\dfrac{5}{56.61}\)
\(\dfrac{1}{4}A=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+........+\dfrac{1}{51}-\dfrac{1}{56}+\dfrac{1}{56}-\dfrac{1}{61}\)
\(\dfrac{1}{4}A=1-\dfrac{1}{61}\)
\(\Rightarrow A=\dfrac{60}{61}:\dfrac{1}{4}\)
\(\Rightarrow A=\dfrac{60}{61}.4\)
\(\Rightarrow A=\dfrac{240}{61}\)
Cậu làm đúng rồi đó!!!