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22 tháng 3 2018
\(Q=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
\(Q=\left(\frac{1}{2}\right).\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)...\left(\frac{99}{100}\right)\)
\(Q=\frac{1}{100}\)
\(P=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(P=\left(\frac{1.3}{1.3}+\frac{1}{1.3}\right)\left(\frac{2.4}{2.4}+\frac{1}{2.4}\right)\left(\frac{3.5}{3.5}+\frac{1}{3.5}\right)...\left(\frac{99.101}{99.101}+\frac{1}{99.101}\right)\)
\(P=\left(\frac{4}{1.3}\right)\left(\frac{9}{2.4}\right)\left(\frac{16}{3.5}\right)...\left(\frac{10000}{99.101}\right)\)
\(P=\left(\frac{2^2}{1.3}\right)\left(\frac{3^2}{2.4}\right)\left(\frac{4^2}{3.5}\right)...\left(\frac{100^2}{99.101}\right)\)
Bạn tự tách ra rồi bạn sẽ ra kết quả như ở dưới
\(P=\frac{201}{100}\)
4 tháng 2 2017
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2.\left(1-\frac{1}{99}\right)\)
\(=2.\frac{98}{99}\)
\(=\frac{196}{99}=1\frac{97}{99}\)
25 tháng 3 2015
Ta có:
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...\frac{9}{98.99}+\frac{9}{99.100}\)
\(=9.\frac{1}{1.2}+9.\frac{1}{2.3}+9.\frac{1}{3.4}+...+9.\frac{1}{98.99}+9.\frac{1}{99.100}\)
\(=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=9.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9.\left(1-\frac{1}{100}\right)\)
\(=9.\frac{99}{100}\)
\(=\frac{9.99}{100}\)
\(=\frac{891}{100}\)
16 tháng 4 2018
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=9.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
5 tháng 3 2017
\(A=\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.100}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{3}-\frac{1}{100}=\frac{97}{300}\)
TN
23 tháng 4 2016
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9\left(1-\frac{1}{100}\right)\)
\(=9\times\frac{99}{100}\)
\(=\frac{891}{100}\)
23 tháng 4 2016
A=9.(1/1.2 +1/2.3 +1/3.4+...+1/98.99 +1/99.100
A=9.(1-1/2+1/2-1/3+1/3-1/4+...+1/98-1/99+1/99-1/100)
A=9.(1-1/100)
A=9.99/100
A=891/100
\(M=\frac{\left(2.4....98.100\right).\left(51.52....99.100\right)}{\left(2.4....98.100\right).\left(1.3....97.99\right).2^{50}}\)
\(M=\frac{\left(2.1\right).\left(2.3\right)....\left(2.49\right).\left(2.50\right).51.52....99.100}{1.2......99.100.2^{50}}\)
\(M=\frac{2^{50}.\left(1.2....99.100\right)}{\left(1.2....99.100\right).2^{50}}\)\(=1\)
Vậy M =1
Chúc bạn học tốt ( -_- )
\(M=\frac{51.52...99.100}{1.3...97.99.2^{50}}\)
\(=\frac{\left(1.2.3...50\right).51.52...99.100}{1.3...97.99.2^{50}.\left(1.2.3...50\right)}\)
\(=\frac{1.2.3...99.100}{1.3...97.99.\left(2.4.6...100\right)}\)
\(=\frac{1.2.3...99.100}{1.2.3...99.100}\)