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222-9+11+12:2*14+14 = ? ( )

## Toán Lớp 6: P=(1/2^2-1) x (1/3^2-1) x … (1/2021^2-1). So sánh P với 1/2

Toán Lớp 6: P=(1/2^2-1) x (1/3^2-1) x … (1/2021^2-1). So sánh P với 1/2

## Comments ( 2 )

1. Xin ctlhn

2. Do có $2020$ cặp số nên
$\begin{array}{l} \left( {\dfrac{1}{{{2^2}}} – 1} \right)\left( {\dfrac{1}{{{3^2}}} – 1} \right)…\left( {\dfrac{1}{{{{2021}^2}}} – 1} \right)\\ A = \left( {1 – \dfrac{1}{{{2^2}}}} \right)\left( {1 – \dfrac{1}{{{3^2}}}} \right)…\left( {1 – \dfrac{1}{{{{2021}^2}}}} \right)\\ A = \left( {1 – \dfrac{1}{2}} \right)\left( {1 + \dfrac{1}{2}} \right)\left( {1 – \dfrac{1}{3}} \right)\left( {1 + \dfrac{1}{3}} \right)…\left( {1 – \dfrac{1}{{2021}}} \right)\left( {1 + \dfrac{1}{{2021}}} \right)\\ A = \dfrac{1}{2}.\dfrac{3}{2}.\dfrac{2}{3}.\dfrac{4}{3}…\dfrac{{2020}}{{2021}}.\dfrac{{2022}}{{2021}}\\ A = \dfrac{{1.2.3.4.2020}}{{2.3.4….2021}}.\dfrac{{3.4.5…2022}}{{2.3.4…2021}}\\ A = \dfrac{1}{{2021}}.\dfrac{{2022}}{2} = \dfrac{{2022}}{{4042}} > \dfrac{{2022}}{{4044}} = \dfrac{1}{2}\\ \Rightarrow A > \dfrac{1}{2} \end{array}$