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Tính giá trị biểu thức \(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{9.9^2}{9.8.100}\)
\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{10^2}{9.11}=\frac{\left(1.2.3.....10\right)^2}{\left(1.2.3.....9\right).\left(3.4.5....9.10.11\right)}=\frac{\left(1.2.3....10\right)^2}{\left(1.2\right)\left(3.4.5.....9\right)^2\left(10.11\right)}=\frac{\left(1.2.10\right)^2}{\left(1.2\right).\left(10.11\right)}=\frac{1.2.10}{11}=\frac{20}{11}\)
\(\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{98.100}{99^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{98.100}{99.99}\)
\(=\frac{1.2.3...98}{2.3.4...99}.\frac{3.4.5...100}{2.3.4...99}\)
\(=\frac{1}{99}.\frac{100}{2}\)
\(=\frac{1}{99}.50=\frac{50}{99}\)
\(A=\frac{2^2}{1.3}\cdot\frac{2^2}{2.4}\cdot\frac{2^2}{3.5}\cdot\frac{2^2}{4.6}\)
\(A=\frac{4}{3}\cdot\frac{1}{2}\cdot\frac{4}{15}\cdot\frac{1}{6}\)
\(A=\frac{4.1.4.1}{3.2.15.6}\)
\(A=\frac{4}{135}\)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}\)
\(=\frac{2.3.4.5}{1.2.3.4}.\frac{2.3.4.5}{3.4.5.6}\)
\(=\frac{5}{1}.\frac{2}{6}\)
\(=\frac{5}{1}.\frac{1}{3}\)
\(=\frac{5}{3}\)
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 = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
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) \(C=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(C=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(C=\frac{3}{2}\left(\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{61-59}{59.61}\right)\)
\(C=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(C=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(C=\frac{3}{2}.\frac{56}{305}=\frac{84}{305}\)
\(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{2013\cdot2015}\right)\)
\(=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot...\cdot\frac{4056196}{2013\cdot2015}\)
\(=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2014\cdot2014\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2013\cdot2015\right)}\)
\(=\frac{\left(2\cdot3\cdot4\cdot...\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2014\right)}{\left(1\cdot2\cdot3\cdot...\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2015\right)}\)
\(=\frac{2014\cdot2}{1\cdot2015}\)
\(=\frac{4028}{2015}\)
\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}=\frac{2.3.4.5}{1.2.3.4}.\frac{2.3.4.5}{3.4.5.6}=5.\frac{1}{3}=\frac{5}{3}\)