Tính \(\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)...\left(1 +\frac{1}{120}\right)\)
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HN
7 tháng 3 2017
\(\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)...\left(1+\dfrac{1}{120}\right)\)
\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...\dfrac{121}{120}=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.\dfrac{11.11}{10.12}\)
\(=\dfrac{2}{1}.\dfrac{11}{12}=\dfrac{11}{6}\)
18 tháng 5 2017
\(\left(1-\frac{1}{3}\right).\left(1-\frac{1}{5}\right).\left(1-\frac{1}{7}\right)...\left(1-\frac{1}{2}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{6}\right).\left(1-\frac{1}{120}\right)\)
\(=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{119}{120}=\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{118}{119}\right).\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{119}{120}\right)\)
\(=\frac{\left(2.4.6...118\right).\left(1.3.5...119\right)}{\left(3.5.7...119\right).\left(2.4.6...120\right)}=\frac{1}{120}\).
Đáp số: \(\frac{1}{120}\).
1 tháng 4 2015
(1+1/3)(1+1/8)(1+1/15)...(1+1/9603)=4/3 . 9/8 . 16/15 ... 9604/9603
= (2.2)/(1.3) . (3.3)/(2.4) . (4.4)/(3.5) ... (98.98)/(97.99)
=(2.2.3.3.4.4...98.98)/(1.3.2.4.3.5...97.99)
=(2.3.4...98)/(1.2.3...97) . (2.3.4..98)/(3.4.5...99)
=98/1 .2/99 =169/99 .
6 tháng 8 2020
\(\frac{5.18-10.27+15.36}{10.36-20.54+30.72}\)
\(=\frac{5.18-10.27+15.36}{5.2.18.2-10.2.27.2+15.2.36.2}\)
\(=\frac{5.18-10.27+15.36}{5.8.2.2-10.27.2.2+15.36.2.2}\)
\(=\frac{1}{2.2-2.2+2.2}\)
\(=\frac{1}{2.2}=\frac{1}{4}\)
\(\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{120}\right)\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{121}{120}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{11.11}{10.12}\)
\(=\frac{2.2.3.3.4.4...11.11}{1.3.2.4.3.5...10.12}\)
\(=\frac{\left(2.3.4...11\right)\left(2.3.4...11\right)}{\left(1.2.3...10\right).\left(3.2.5...12\right)}\)
\(=\frac{11.2}{1.12}\)
\(=\frac{11}{6}\)