Tính:
\(\dfrac{15}{11.14} +\dfrac{15}{14.17}+\dfrac{15}{17.20}+...+\dfrac{15}{72.75}\)
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\(\dfrac{15}{11.14}+\dfrac{15}{14.17}+\dfrac{15}{17.20}+...+\dfrac{15}{68.71}\)
\(=5\left(\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}+...+\dfrac{1}{68}-\dfrac{1}{71}\right)\)
\(=5\left(\dfrac{1}{11}-\dfrac{1}{71}\right)\)
\(=5.\dfrac{60}{781}\)
\(=\dfrac{300}{781}\)
theo bài ra ta có:
\(E=\dfrac{15}{11.14}+\dfrac{15}{14.17}+\dfrac{15}{17.20}+...+\dfrac{15}{74.77}\\ \Rightarrow\dfrac{1}{5}E=\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+...+\dfrac{3}{74.77}\\ \dfrac{1}{5}E=\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}+...+\dfrac{1}{74}-\dfrac{1}{77}\\ \dfrac{1}{5}E=\dfrac{1}{11}-\dfrac{1}{77}\\ \dfrac{1}{5}E=\dfrac{7}{77}-\dfrac{1}{77}=\dfrac{6}{77}\\ \Rightarrow E=\dfrac{6}{77}.5\\ E=\dfrac{30}{77}\)
5 .\((\)\(\dfrac{3}{11.14}+\dfrac{3}{14.17}+...+\dfrac{3}{74.77}\))
= 5. (\(\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{74}-\dfrac{1}{77}\))
= 5.(\(\dfrac{1}{11}-\dfrac{1}{77}\))
= 5. \(\dfrac{6}{77}\)
= \(\dfrac{30}{77}\)
\(\frac{3}{15}\cdot G=\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+...+\frac{3}{68\cdot71}\)
\(\frac{3}{15}\cdot G=\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{68}-\frac{1}{71}\)
\(\frac{3}{15}\cdot G=\frac{1}{11}-\frac{1}{71}\)
\(G=\frac{60}{781}\cdot\frac{15}{3}\)
\(G=\frac{300}{781}\)
ta có :\(\frac{3}{15}G=\left(\frac{15}{11.14}+\frac{15}{14.17}+...+\frac{15}{68.71}\right)\)
\(\frac{3}{15}G=\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{68.71}\)
\(\frac{3}{15}G=\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{68}-\frac{1}{71}\)
\(\frac{3}{15}G=\frac{1}{11}-\frac{1}{71}=\frac{71}{781}-\frac{11}{781}=\frac{60}{781}\)
\(=>G=\frac{60}{781}:\frac{3}{15}=\frac{900}{2343}\)
vậy G =900/2343
\(A=\dfrac{14}{8.11}+\dfrac{14}{11.14}+\dfrac{14}{14.17}+.....+\dfrac{14}{197.200}\)
\(A=\dfrac{14}{3}\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\dfrac{24}{200}=\dfrac{28}{25}\)
\(B=\dfrac{7}{15}+\dfrac{7}{35}+\dfrac{7}{63}+...+\dfrac{7}{399}\)
\(B=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+.....\dfrac{7}{19.21}\)
\(B=\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\dfrac{6}{21}=1\)
`3x-15/(5*8)-15/(8*11)-15/(11*14)-...-15/(47*50)=2 1/10`
`3x-(15/(5*8)+15/(8*11)+15/(11*14)+...+15/(47*50))=21/10`
`3x-5(3/(5*8)+3/(8*11)+3/(11*14)+...+3/(47*50))=21/10`
`3x-5(1/5-1/8+1/8-1/11+1/11-1/14+...+1/47-1/50)=21/10`
`3x-5(1/5-1/50)=21/10`
`3x-5*9/50=21/10`
`3x-9/10=21/10`
`3x=21/10+9/10`
`3x=3`
`x=1`
\(\frac{15}{11.14}+\frac{15}{14.17}+\frac{15}{17.20}+.......+\frac{15}{74.77}\)
\(=5\left(\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}+.......+\frac{3}{74.77}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}+.....+\frac{1}{74}-\frac{1}{77}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{77}\right)\)
\(=5\left(\frac{7}{77}-\frac{1}{77}\right)\)
\(=5.\frac{6}{77}\)
\(=\frac{30}{77}\)
\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\)
= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\)
\(=\dfrac{1}{2}-\dfrac{1}{17}\)
\(=\dfrac{15}{34}\)
Vì \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)
\(\frac{15}{11.14}+\frac{15}{14.17}+\frac{15}{17.20}+...+\frac{15}{72.75}\)
\(=5\left(\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}+...+\frac{3}{72.75}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}+...+\frac{1}{72}-\frac{1}{75}\right)\)\(=5\left(\frac{1}{11}-\frac{1}{75}\right)\)
\(=\frac{64}{165}\)
pài này gần giống pài troq v15