Tích trên có thừa số 1 - 253 = -252 còn các thừa số kia trong tích đều dương. Vậy tích trên âm.

Mà 2/5 dương nên đpcm

de yeu cau gi Phạm Khánh Linh

(1−13)(1−16)...(1−1253)

=23⋅56⋅...⋅252253=46⋅1012⋅...⋅504506

=1⋅42⋅3⋅2⋅53⋅4⋅...⋅21⋅2422⋅23

=1⋅2⋅3⋅42⋅52⋅...⋅212⋅22⋅23⋅242⋅32⋅42⋅...⋅222⋅23=1⋅243⋅22=2466<25

Xin người.

\(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)...\left(1-\dfrac{1}{253}\right)\\=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{252}{253}\\ =\dfrac{4}{6}\cdot\dfrac{10}{12}\cdot...\cdot\dfrac{504}{506}\\ =\dfrac{1\cdot4}{2\cdot3}\cdot\dfrac{2\cdot5}{3\cdot4}\cdot...\cdot\dfrac{21\cdot24}{22\cdot23}\\ =\dfrac{1\cdot2\cdot3\cdot4^2\cdot5^2\cdot...\cdot21^2\cdot22\cdot23\cdot24}{2\cdot3^2\cdot4^2\cdot...\cdot22^2\cdot23}\\ =\dfrac{1\cdot24}{3\cdot22}=\dfrac{24}{66}< \dfrac{2}{5}\)

Bài 1: 

Ta có: \(\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{6}-1\right)\left(\dfrac{1}{10}-1\right)\cdot...\cdot\left(\dfrac{1}{45}-1\right)\)

\(=\dfrac{-2}{3}\cdot\dfrac{-5}{6}\cdot\dfrac{-9}{10}\cdot...\cdot\dfrac{-44}{45}\)

\(=\dfrac{-2}{3}\cdot\dfrac{-5}{6}\cdot\dfrac{-9}{10}\cdot\dfrac{-14}{15}\cdot\dfrac{-20}{21}\cdot\dfrac{-27}{28}\cdot\dfrac{-35}{36}\cdot\dfrac{-44}{45}\)

\(=\dfrac{11}{27}\)

Câu 2: 

B=1+1/2+1/3+....+1/2010

 =(1+1/2010)+(1/2+1/2009)+(1/3+1/2008)+...(1/1005+1/1006)

 = 2011/2010+2011/2.2009+2011/3.2008+...+2011/1005.1006

 =2011.(1/2010+.....1/1005.1006)

Vậy B có tử số chia hết cho 2011 (đpcm).

Câu 3:

 \(P=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}....\dfrac{98}{99}\\ P< \dfrac{3}{4}.\dfrac{5}{6}.\dfrac{6}{7}....\dfrac{99}{100}\\ P^2< \dfrac{2}{100}\)

Mà

 \(\dfrac{2}{100}=\dfrac{1}{50}< \dfrac{1}{49}\\ \Rightarrow P< \dfrac{1}{7}\)

B

B

B. \(\dfrac{8}{11}\)