Trả lời:
\(P=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2021^2}-1\right)\)
\(=\frac{1-2^2}{2^2}\cdot\frac{1-3^2}{3^2}\cdot\frac{1-4^2}{4^2}\cdot...\cdot\frac{1-2021^2}{2021^2}\)
\(=\frac{-3}{2^2}\cdot\frac{-8}{3^2}\cdot\frac{-15}{4^2}\cdot...\cdot\frac{-4084440}{2021^2}\)
\(=\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot...\cdot\frac{4084440}{2021^2}\) ( vì tích trên có 2020 thừa số, mà tích của 2020 thừa số âm là số dương )
\(=\frac{3\cdot8\cdot15\cdot...\cdot4084440}{2^2\cdot3^2\cdot4^2\cdot...\cdot2021^2}\)
\(=\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot2020\cdot2022}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot2021\cdot2021}\)
\(=\frac{\left(1\cdot2\cdot3\cdot...\cdot2020\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot2022\right)}{\left(2\cdot3\cdot4\cdot...\cdot2021\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot2021\right)}\)
\(=\frac{1\cdot2022}{2021\cdot2}=\frac{1011}{2021}>\frac{1011}{2022}=\frac{1}{2}\)
Vậy \(P>\frac{1}{2}\)
Kết luận B<1/2
\(P=\left(\frac{1}{2^2}-1\right)\times\left(\frac{1}{3^2}-1\right)\times...\times\left(\frac{1}{2021^2}-1\right)\)
\(=\left(1-\frac{1}{2^2}\right)\times\left(1-\frac{1}{3^2}\right)\times...\times\left(1-\frac{1}{2021^2}\right)\)
\(=\frac{2^2-1}{2^2}\times\frac{3^2-1}{3^2}\times...\times\frac{2021^2-1}{2021^2}\)
\(=\frac{1\times3}{2^2}\times\frac{2\times4}{3^2}\times\frac{3\times5}{4^2}\times...\times\frac{2020\times2022}{2021^2}\)
\(=\frac{\left(1\times2\times3\times...\times2020\right)\times\left(3\times4\times5\times...\times2022\right)}{\left(2\times3\times4\times...\times2021\right)\times\left(2\times3\times4\times...\times2021\right)}\)
\(=\frac{1\times2022}{2021\times2}=\frac{2022}{2021\times2}>\frac{2021}{2021\times2}=\frac{1}{2}\)
Ta có 1 - a2 = 1 - a + a - a2 = 1 - a + a(1 - a) = (1 - a)(a + 1)
Khi đó P = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2021^2}-1\right)\)
= \(\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}....\frac{1-2021^2}{2021^2}\)
= \(\frac{\left(1-2\right)\left(1+2\right).\left(1-3\right)\left(1+3\right).\left(1-4\right)\left(1+4\right)....\left(1-2021\right)\left(1+2021\right)}{2^2.3^2.4^2....2021^2}\)
\(=\frac{-1.3.\left(-2\right).4.\left(-3\right).5...\left(-2020\right).2022}{2^2.3^2.4^2...2021^2}\)
\(=-\frac{\left(1.2.3...2020\right).\left(3.4.5...2022\right)}{\left(2.3.4....2021\right).\left(2.3.4...2021\right)}=-\frac{1.2022}{2021.2}=-\frac{1011}{2021}\)
P=1−122.1−132.1−142.....1−1502P=1−122.1−132.1−142.....1−1502
P=022.032.....0502P=022.032.....0502
P=0.0.0.....0P=0.0.0.....0
P=0P=0
−>P<12−>P<12
