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NT
1
IH
22 tháng 7 2015
Làm lại đề cho:
\(\left(\frac{1}{4}-1\right)\cdot\left(\frac{1}{9}-1\right)\cdot\left(\frac{1}{16}-1\right)...\left(\frac{1}{81}-1\right)\cdot\left(\frac{1}{100}-1\right)\)
Tính nhẩm
MT
9 tháng 8 2016
\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{81}-1\right)\left(\frac{1}{100}-1\right)\)
\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}....\frac{-80}{81}.\frac{-99}{100}\)
\(=\left[\left(-1\right).\left(-1\right)...\left(-1\right)\left(9\text{số (-1)}\right)\right].\frac{3}{4}.\frac{8}{9}....\frac{99}{100}\)
\(=\left(-1\right).\frac{1.3}{2.2}.\frac{2.4}{3.3}....\frac{9.11}{10.10}\)
\(=-\frac{1.11}{2.10}=-\frac{11}{10}\)
14 tháng 9 2016
\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{81}-1\right)\left(\frac{1}{100}-1\right)\)
\(=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....\frac{-80}{81}.\frac{-99}{100}\)
\(=\left[\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right).\left(-1\right)\right].\frac{3}{4}.\frac{8}{9}.....\frac{99}{100}\)
\(=\left(-1\right).\frac{1.3}{2.2}.\frac{2.4}{3.3}....\frac{9.11}{10.10}\)
\(=-\frac{1.11}{2.10}=-\frac{11}{10}\)
NT
0
S
12 tháng 10 2019
\(A=\left(\frac{1-2^2}{2^2}\right)\left(\frac{1-3^2}{3^2}\right)....\left(\frac{1-10^2}{10^2}\right)\)
\(A=\frac{\left(1+2\right)\left(1-2\right)}{2^2}.\frac{\left(1-3\right)\left(1+3\right)}{3^2}.......\frac{\left(1-10\right)\left(1+10\right)}{10^2}\)
\(A=\frac{3.\left(-1\right)}{2^2}.\frac{\left(-2\right).4}{3^2}........\frac{\left(-9\right).11}{10^2}=-\left(\frac{1.3}{2^2}.\frac{2.4}{3^2}....\frac{9.11}{10^2}\right)\)
\(=-\left(\frac{1.2....9}{2.3....10}.\frac{3.4....11}{2.3.4...10}\right)=-\left(\frac{1}{10}.\frac{11}{2}\right)=\frac{-11}{20}\)
23 tháng 9 2021
\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{81}\right)\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{99}{100}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}...\dfrac{9.11}{10.10}=\left(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{9}{10}\right).\left(\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{11}{10}\right)=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)
23 tháng 9 2021
\(B=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{9}\right)\left(1+\dfrac{1}{9}\right)\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\\ B=\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{8}{9}\cdot\dfrac{9}{10}\right)\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{10}{9}\cdot\dfrac{11}{10}\right)\\ B=\dfrac{1}{10}\cdot\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)
\(=\frac{1}{1.3}.\frac{1}{2.4}...\frac{1}{9.11}=\frac{1}{1.2.3^2...9^2.10.11}\)