Question

[Trương Tùng Dương]

Cho A= \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right).......\left(\frac{1}{2018^2}-1\right)\)

B= \(-\frac{1}{2}\)

So sánh A và B

Answer 1

[Nguyễn Phương Uyên]

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2018^2}-1\right)\)

\(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{2018^2}\right)\)

\(-A=\frac{3}{2\cdot2}\cdot\frac{8}{3\cdot3}\cdot...\cdot\frac{4072323}{2018\cdot2018}\)

\(-A=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\cdot...\cdot\left(2017\cdot2019\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\cdot...\cdot\left(2018\cdot2018\right)}\)

\(-A=\frac{\left(1\cdot2\cdot...\cdot2017\right)\left(3\cdot4\cdot...\cdot2019\right)}{\left(2\cdot3\cdot...\cdot2018\right)\left(2\cdot3\cdot...\cdot2018\right)}\)

\(-A=\frac{1\cdot2019}{2018\cdot2}\)

\(-A=\frac{2019}{4036}\)

\(A=-\frac{2019}{4036}< -\frac{1}{2}\)

Answer 2

[SKT Khan]

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2018^2}-1\right)\)

\(\Rightarrow A=\left(\frac{1}{2^2}-\frac{2^2}{2^2}\right)\left(\frac{1}{3^2}-\frac{3^2}{3^2}\right)...\left(\frac{1}{2018}-\frac{2018^2}{2018^2}\right)\)

\(\Rightarrow A=\frac{-3}{2^2}.\frac{-8}{3^2}....\frac{-4072323}{2018^2}\)

\(\Rightarrow\frac{-\left(3.8.....4072323\right)}{\left(2.3.4...2018\right).\left(2.3.4..2018\right)}\)

\(\Rightarrow A=\frac{-\left(1.3.2.4....2017.2019\right)}{\left(2.3.4...2018\right)\left(2.3.4..2018\right)}\)

\(\Rightarrow A=\frac{-\left(\left(1.2.3...2017\right).\left(3.4.5..2019\right)\right)}{\left(2.3...2018\right)\left(2.3.4..2018\right)}\)

\(\Rightarrow A=\frac{-2019}{2018.2}< -\frac{2018}{2018.2}=\frac{-1}{2}\)

\(\Rightarrow A< \frac{-1}{2}\)

P/s: mk ko copy baì của bn uyên đâu nha