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\(A=\frac{2^2}{1.3}\cdot\frac{2^2}{2.4}\cdot\frac{2^2}{3.5}\cdot\frac{2^2}{4.6}\)
\(A=\frac{4}{3}\cdot\frac{1}{2}\cdot\frac{4}{15}\cdot\frac{1}{6}\)
\(A=\frac{4.1.4.1}{3.2.15.6}\)
\(A=\frac{4}{135}\)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}\)
\(=\frac{2.3.4.5}{1.2.3.4}.\frac{2.3.4.5}{3.4.5.6}\)
\(=\frac{5}{1}.\frac{2}{6}\)
\(=\frac{5}{1}.\frac{1}{3}\)
\(=\frac{5}{3}\)
C=2.2.3.3.4.4.5.5/1.3.2.4.3.5.4.6
C=(2.3.4.5).(2.3.4.5)/(1.2.3.4).(3.4.5.6)
C=2.5/6
C=5/3
C=2^2/1.3.3^2/2.4.5^2/3.5.4^2/4.6
C= 2.2.3.3.5.5.4.4/1.3.2.4.3.5.4.6
C=(2.3.4.5).(2.3.4.5)/(1.2.3.4).(3.4.5.6)
C=5.2/6
C=5/3
a, \(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{999^2}{998.1000}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{999.999}{998.1000}\)
\(=\frac{2.3.4...999}{1.2.3...998}.\frac{2.3.4...999}{3.4.5...1000}\)
\(=\frac{999}{1}.\frac{2}{1000}\)
\(=\frac{999.2}{1000.1}=\frac{999.2}{500.2.1}\)
\(=\frac{999}{500}\)
Vậy \(A=\frac{999}{500}\)
chúc bạn học giỏi
=\(\frac{1.3.2.4.3.5...2016.2018}{2.2.3.3.4.4...2017.2017}\)
Ta tách thành hai dãy trên cả mẫu và tử và được \(\frac{\left(1.2.3...2016\right).\left(3.4.5...2018\right)}{\left(2.3.4...2017\right).\left(2.3.4...2017\right)}\)
Giờ thì sẽ rút gọn được kết quả=\(\frac{2018}{2017.2}=\frac{1009}{2017}\)
bạn phải cho ra 2 số cuối thì mới làm đc nha có 1 s
ố cuối ko làm đc đâu
A= 1-1/2 + 1-1/3 + 1/2-1/5 + 1/3-1/8+ 1/5-1/13+1/8- 1/21 +....+ 1/610- 1/1597
A= 1/610
1)\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
2)\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2\times\frac{502}{1005}\)
\(=\frac{1004}{1005}\)
tự làm tiếp nhé
1.= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
= \(1-\frac{1}{101}\) = \(\frac{100}{101}\)
2.= \(2\cdot\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)
= \(2\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
= \(2\cdot\left(\frac{1}{2}-\frac{1}{2010}\right)\) = \(2\cdot\frac{502}{1005}\) = \(\frac{1004}{1005}\)
\(I=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{999^2}{998.1000}\)
\(I=\frac{2^2.3^2.4^2...999^2}{2.3^2.4^2...998^2.1000}\)
\(I=\frac{2}{1000}=\frac{1}{500}\)
\(=\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\frac{4\cdot4}{3\cdot5}\cdot...\cdot\frac{999\cdot999}{998\cdot1000}\)
\(=\frac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot999\cdot999}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot998\cdot1000}\)
\(=\frac{2\cdot3\cdot4\cdot...\cdot999}{1\cdot2\cdot3\cdot...\cdot998}\cdot\frac{2\cdot3\cdot4\cdot...\cdot999}{3\cdot4\cdot5\cdot...\cdot1000}\)
\(=\frac{999}{1}\cdot\frac{2}{1000}\)
\(=\frac{999}{500}\)
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
B = \(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}=\frac{\left(2.3.4.5\right).\left(2.3.4.5\right)}{\left(1.2.3.4\right).\left(3.4.5.6\right)}=\frac{5.2}{1.6}=\frac{5}{3}\)
C = \(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{3}{2}.\frac{56}{305}=\frac{74}{305}\)
Bài làm:
1) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}=\frac{49}{50}\)
2) \(B=\frac{2^2.3^2.4^2.5^2}{1.2.3^2.4^2.5.6}=\frac{2.5}{6}=\frac{5}{3}\)
3) \(C=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(C=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(C=\frac{3}{2}\left(\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{61-59}{59.61}\right)\)
\(C=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(C=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(C=\frac{3}{2}.\frac{56}{305}=\frac{84}{305}\)
\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}=\frac{2.3.4.5}{1.2.3.4}.\frac{2.3.4.5}{3.4.5.6}=5.\frac{1}{3}=\frac{5}{3}\)
233556)₫₫₫₫)7789900”99866534233579976543322234555688999887654556):/223477790”987776655655555555554324
=16.83313426
\(\frac{2^2}{1\cdot3}+\frac{3^2}{2\cdot4}+\frac{4^2}{3\cdot5}+\frac{5^2}{4\cdot6}=\frac43+\frac98+\frac{16}{15}+\frac{25}{24}=\frac{137}{30}\)
Ta có: \(\frac{2^2}{1\cdot3}+\frac{3^2}{2\cdot4}+\frac{4^2}{3\cdot5}+\frac{5^2}{4\cdot6}\)
\(=\frac43+\frac98+\frac{16}{15}+\frac{25}{24}\)
\(=1+\frac13+1+\frac18+1+\frac{1}{15}+1+\frac{1}{24}\)
\(=4+\left(\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\frac{1}{3\cdot5}+\frac{1}{4\cdot6}\right)\)
\(=4+\frac12\left(\frac{2}{1\cdot3}+\frac{2}{2\cdot4}+\frac{2}{3\cdot5}+\frac{2}{4\cdot6}\right)\)
\(=4+\frac12\left(1-\frac13+\frac12-\frac14+\frac13-\frac15+\frac14-\frac16\right)\)
\(=4+\frac12\left(1+\frac12-\frac15-\frac16\right)=4+\frac12\left(\frac32-\frac15-\frac16\right)\)
\(=4+\frac12\left(\frac{45}{30}-\frac{6}{30}-\frac{5}{30}\right)=4+\frac12\cdot\frac{34}{30}=4+\frac{17}{30}=\frac{120+17}{30}=\frac{137}{30}\)