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bài 1b)
\(8\frac{1}{14}-6\frac37\)
C1:\(\frac{113}{14}-\frac{45}{7}\) =\(\frac{113}{14}-\frac{90}{14}=\frac{23}{14}\)
C2:\(8\frac{1}{14}-6\frac37=\left(8-6\right)+\left(\frac{1}{14}-\frac37\right)=2+\left(\frac{1}{14}-\frac{6}{14}\right)\)
\(=2+\frac{-5}{14}=\frac{28}{14}-\frac{5}{14}=\frac{23}{14}\)
bài 1 c)\(7-3\frac67\)
C1:\(\) \(7-3\frac67=7-\frac{27}{7}=\frac{49}{7}-\frac{27}{7}=\frac{22}{7}\)
C2:\(7-3\frac67=\left(7-3\right)-\frac67=4-\frac67=\frac{28}{7}-\frac67=\frac{22}{7}\)
\(\text{ta có:}\frac{6}{a\left(a+7\right)}+1=\frac{\left(a+1\right)\left(a+6\right)}{a\left(a+7\right)}\text{ do đó:}A=\frac{2.7}{1.8}.\frac{3.8}{2.9}.....\frac{101.106}{100.107}\)
\(=\frac{2.3...101.\left(7.8....106\right)}{1....101.\left(8.9.....107\right)}=\frac{7}{107}\)
\(T=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{11}\right)\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{8}\right)\left(1-\frac{1}{10}\right)\)
\(T=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.\frac{8}{9}.\frac{10}{11}.\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}.\frac{9}{10}=\frac{\left(2.4.6.8.10\right).\left(1.3.5.7.9\right)}{\left(3.5.7.9.11\right).\left(2.4.6.8.10\right)}=\frac{1}{11}\)
Đáp số: T=1/11