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a, \(M=\frac{3}{2}\cdot\frac{4}{3}\cdot\cdot\cdot\cdot\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{3.4...2019}{2.3...2018}=\frac{2019}{2}\)
b, c cùng 1 câu phải k
ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{2018}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=B\)
\(\Rightarrow\frac{A}{B}=1\Rightarrow\left(\frac{A}{B}\right)^{2018}=1^{2018}=1\)
A,\(M=\frac{3}{2}\cdot\frac{4}{3}....\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{4\cdot3...2019}{2\cdot3...2018}=\frac{2019}{2}\)
NHA
HỌC TỐT
\(A=\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+\frac{6}{2}+..........+\frac{2019}{2}=\frac{3+4+5+..............+2019}{2}.\)
ta có 3+4+5+......+2019=(3+2019)2016:2=2038176
=>\(\frac{2038176}{2}=1019088\)
\(\Rightarrow C=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)....\left(1-\frac{1}{1+2++...+2017}\right)\)
\(\Rightarrow C=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)....\left(1-\frac{1+2+...+2017-1}{1+2+3+...+2017}\right)\)
\(\Rightarrow C=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)....\left(\frac{2017.2018:2-1}{2017.2018:2}\right)\)
\(\Rightarrow C=\frac{2}{3}.\frac{5}{6}...\frac{2017.2018:2-1}{2017.2018:2}\)
\(\Rightarrow C=\frac{4}{6}.\frac{10}{12}....\frac{2017.2018-1}{2017.2018}\)
\(\Rightarrow C=\frac{1.4}{2.3}.\frac{2.5}{3.4}....\frac{2016.2019}{2017.2018}\)
\(\Rightarrow C=\frac{1.2....2016}{2.3.....2018}.\frac{4.5....2019}{3.4....2017}\)
\(\Rightarrow C=\frac{1}{2017.2018}.\frac{2018.2019}{3}\)
\(\Rightarrow C=\frac{673}{2017}\)
đầu bài mình đặt là A mà giải lại là C =.= nhưng mà ko sao vì bạn làm đúng rồi. cảm ơn An Nguyễn nhé <3
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
B=1-1/2017
B=2016/2017
(Vì ta đưa bài toán về cách khử đi .VD:1/2+1/4+1/8=1-1/2+1/2-1/4+1/4-1/8 =1-1/2+1/2-1/4+1/4-1/8=1-1/8=7/8)
(Tui mới học lớp 5 thôi)
Giải
B=\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)+...\left(1-\frac{1}{1+2+3+...+2017}\right)\)
=\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)...\left(1-\frac{1}{\left(1+2017\right).2017:2}\right)\)
=\(\frac{2}{3}.\frac{5}{6}...\frac{2018.2017:2-1}{2018.2017:2}\)
=\(\frac{4}{6}.\frac{10}{12}...\frac{\left(2018.2017:2-1\right).2}{2018.2017}\)
=\(\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{2016.2019}{2017.2018}\)
=\(\frac{1.2...2016}{3.4....2018}.\frac{4.5...2019}{2.3...2017}\)
=\(\frac{2}{2017.2018}.\frac{2018.2019}{2.3}\)
=\(\frac{2019}{2017.3}=\frac{2019}{6051}=\frac{613}{2017}\)
\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{2016}-1\right)\left(\frac{1}{2017}-1\right)\)
\(=\frac{-1}{2}.\frac{-2}{3}...\frac{-2015}{2016}.\frac{-2016}{2017}\)
\(=\frac{1.2...2015.2016}{2.3...2016.2017}\) ( tử số có chẵn số hạng )
\(=\frac{1}{2017}\)
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{2017}\left(1+2+...+2017\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+....+\frac{1}{2017}.\frac{2017\left(2017+1\right)}{2}\)
\(=1+\frac{2.3}{2.2}+\frac{3.4}{3.2}+....+\frac{2017.2018}{2017.2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{2018}{2}\)
\(=\frac{2+3+4+...+2018}{2}\)
\(=\frac{\frac{2018\left(2018+1\right)}{2}-1}{2}\)
\(=1018585\)
Suy ra A=1+1.5+2+....+1009=1 013 532.5