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Câu hỏi của Trần Minh Hưng - Toán lớp | Học trực tuyến
Mik lười quá bạn tham khảo câu 3 tại đây nhé:
Câu hỏi của nguyen linh nhi - Toán lớp 6 - Học toán với OnlineMath
\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\)
\(2S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)
\(2S=\frac{1}{2}-\frac{1}{38\cdot39}\)
\(S=\frac{1}{4}-\frac{1}{2\cdot38\cdot39}< \frac{1}{4}\)
Ta có : \(\frac{1}{m}=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{59.60}=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{60}\right)=\frac{19}{30}\)
\(\Rightarrow m=\frac{30}{19}>\frac{2}{3}\)
\(Tac\text{ó}:\frac{1}{m}=\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{59.60}=2\left(\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{59}-\frac{1}{60}\right)\)
\(=>2\left(\frac{1}{3}-\frac{1}{60}\right)=\frac{19}{30}\\ =>m=\frac{30}{19}>\frac{2}{3}\)
\(\frac{1}{M}=\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+...+\frac{1}{1+2+3+...+59}\)
\(\frac{1}{M}=\frac{1}{3\left(1+3\right):2}+\frac{1}{4\left(1+4\right):2}+\frac{1}{5\left(1+5\right):2}+...+\frac{1}{59\left(1+59\right):2}\)
\(\frac{1}{M}=\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{59.60}\)
\(\frac{1}{M}=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\right)\)
\(\frac{1}{M}=2\left(\frac{1}{3}-\frac{1}{60}\right)\)
\(\frac{1}{M}=\frac{1}{2}.\frac{19}{60}\)
\(\frac{1}{M}=\frac{19}{120}\)
\(M=\frac{120}{19}>\frac{2}{3}\left(đpcm\right)\)
\(M=\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+59}\)
\(M=\frac{1}{\frac{3.\left(3+1\right)}{2}}+\frac{1}{\frac{4.\left(4+1\right)}{2}}+\frac{1}{\frac{5.\left(5+1\right)}{2}}+...+\frac{1}{\frac{59.\left(59+1\right)}{2}}\)
\(M=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+\frac{1}{\frac{5.6}{2}}+...+\frac{1}{\frac{59.60}{2}}\)
\(M=\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{59.60}\)
\(M=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{59.60}\right)\)
\(M=2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\right)\)
\(M=2.\left(\frac{1}{3}-\frac{1}{60}\right)\)
\(M< 2.\frac{1}{3}\)
\(M< \frac{2}{3}\)
M=[ 1+1/2018 +1/2 +1/2017 +1/3 +1/2016 +........+1/1009 +1/1010] .2.3.4...2018
M=[2019/2018 =2019/2.2017 +2019/3.2016 +....+2019/1009.1010].2.3.....2018
M.=2019.[1/2018 +1/2.2017 +.....+1/1009.1010] .2.3....2018 chia het cho 2019
suy ra M chia het cho2019
vay M chia het cho2019
\(M=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{4005}\)
\(\frac{M}{2}=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{8010}\)
\(\frac{M}{2}=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{89x90}\)
\(\frac{M}{2}=\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{90-89}{89.90}\)
\(\frac{M}{2}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{89}-\frac{1}{90}=\frac{1}{3}-\frac{1}{90}\)
\(M=\frac{2}{3}-\frac{2}{90}< \frac{2}{3}\)