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\(\left(1-\dfrac{28}{10}\right)\left(1-\dfrac{52}{22}\right)\left(1-\dfrac{80}{36}\right)...\left(1-\dfrac{21808}{10900}\right)\)
\(=\left(1-\dfrac{2.10+8}{10}\right)\left(1-\dfrac{2.22+8}{22}\right)\left(1-\dfrac{2.36+8}{36}\right)...\left(1-\dfrac{2.10900+8}{10900}\right)\)
\(=\left(1-2-\dfrac{8}{10}\right)\left(1-2-\dfrac{8}{22}\right)\left(1-2-\dfrac{8}{36}\right)...\left(1-2-\dfrac{8}{10900}\right)\)
\(=\left(-1-\dfrac{8}{1.10}\right)\left(-1-\dfrac{8}{2.11}\right)\left(-1-\dfrac{8}{3.12}\right)...\left(-1-\dfrac{8}{100.109}\right)\)
\(=\left(\dfrac{-18}{1.10}\right)\left(\dfrac{-30}{2.11}\right)\left(\dfrac{-44}{3.12}\right)...\left(\dfrac{-10908}{100.109}\right)\)
\(=\left(\dfrac{-2.9}{1.10}\right)\left(\dfrac{-3.10}{2.11}\right)\left(\dfrac{-4.11}{3.12}\right)...\left(\dfrac{-101.108}{100.109}\right)\)
\(=\dfrac{\left(-2\right)\left(-3\right)\left(-4\right)...\left(-101\right)}{1.2.3...100}.\dfrac{9.10.11...108}{10.11.12...109}\) (1)
\(=\dfrac{101}{1}.\dfrac{9}{109}=\dfrac{909}{109}\)
Do ở (1) có \(-2-\left(-101\right)+1=100\) nhân tử (số nhân tử là số chẵn) mang dấu âm nên kết quả sẽ mang dấu dương
Làm luôn nhé
\(A=\frac{3}{8}+\frac{1}{5}+\frac{5}{6}>\frac{1}{6}+\frac{5}{6}=1\)
\(A=\frac{3}{8}+\frac{1}{5}+\frac{5}{6}< \frac{3}{8}+\frac{1}{4}+\frac{5}{4}=\frac{3}{8}+\frac{2}{8}+\frac{10}{8}=\frac{15}{8}< \frac{16}{8}=2\)
Vậy 1<A<2
\(B=\frac{5}{11}+\frac{5}{12}+\frac{5}{13}+\frac{5}{14}>\frac{5}{14}.4=\frac{10}{7}>1\)
\(B=\frac{5}{11}+\frac{5}{12}+\frac{5}{13}+\frac{5}{14}< \frac{5}{10}.4=2\)
Vậy 1<B<2
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