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\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{13\cdot14}+\frac{1}{14\cdot15}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}\)
\(=\frac{1}{2}-\frac{1}{15}\)
\(=\frac{13}{30}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{12.13}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{12}-\frac{1}{13}\)
\(=1-\frac{1}{13}\)
\(=\frac{12}{13}\)
1) A=7/10.11+7/11.12+7/12.13+...+7/69.70
A=7.(1/10.11+1/11.12+1/12.13+...+1/69.70)
A= 7.(1/10-1/11+1/11-1/12+1/12-1/13+...+1/69-1/70)
A= 7.(1/10-1/70)
A=7.3/35=3/5
2)B=1/25.27+1/27.29+1/29.31+...+1/73.1/75
B=1/25-1/27+1/27-1/29+1/29-1/31+...+1/73-1/75
B=1/25-1/75=2/75
A = 7/ 10.11 + 7/ 11.12 + 7/ 12.13 + .... + 7/69.70
(1/7).A=1/10.11+1/11.12+...+1/69.70
=1/10-1/11+1/11-1/12+...+1/69-1/70
=1/10-1/70=3/35
=>A=7.(3/35)
=3/5
2 ) B = 1/ 25.27 + 1/ 27.29 + 1/29.31+ ......+ 1/ 73.75
=>(1/2).B=2/25.27+...+2.73.75
=1/25-1/27+...+1/73-1/75
=1/25-1/75
=2/75
=>B=4/75
\(A=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=\dfrac{7.60}{700}=\dfrac{420}{700}=\dfrac{3}{5}\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{75}\right)=\dfrac{1}{75}\)
\(B=\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+...+\frac{5}{19.20}\)
\(B=5.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{19.20}\right)\)
\(B=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(B=5.\left(\frac{1}{10}-\frac{1}{20}\right)\)
\(B=5.\frac{1}{20}=\frac{1}{4}\)
\(C=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\)
\(4C=4.\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(4C=\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}\)
\(4C=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\)
\(4C=\frac{1}{5}-\frac{1}{25}=\frac{4}{25}\)
\(C=\frac{4}{25}:4=\frac{1}{25}\)
b) Ta có: \(B=\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+\dfrac{1}{12\cdot13}+\dfrac{1}{13\cdot14}\)
\(=\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}\)
\(=\dfrac{1}{10}-\dfrac{1}{14}\)
\(=\dfrac{14}{140}-\dfrac{10}{140}\)
\(=\dfrac{4}{140}=\dfrac{1}{35}\)