Question

Cho: \(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)

So sánh A với \(\frac{-1}{2}\)

Answer 1

Ta có: \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)

\(=\left(-\frac{1.3}{2.2}\right).\left(-\frac{2.4}{3.3}\right)...\left(-\frac{99.101}{100.100}\right)\)

\(=-\frac{1}{2}.\frac{101}{100}=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)

Vậy \(A< -\frac{1}{2}\)

Answer 2

\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)...\left(\frac{1}{10000}-1\right)\)

\(=\frac{-3}{4}\cdot\frac{-8}{9}\cdot\frac{-15}{16}\cdot...\cdot\frac{-9999}{10000}\)

\(=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot...\cdot\frac{-99\cdot111}{100.100}\)

\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot...\cdot\frac{99\cdot111}{100\cdot100}\)

\(=\frac{\left(1\cdot2\cdot3\cdot4\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot6\cdot...\cdot111\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot100\right)^2}\)

\(=\frac{101}{2\cdot100}\)

\(=\frac{101}{200}>\frac{1}{2}\)

Answer 3

Câu trả lời của bạn nên rút gọn lại nha !

~~~~CHÚC CÁC BẠN HỌC TỐT~~~~~ 

Answer 4

\(A=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}...\frac{-9999}{10000}\)

\(A=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.\frac{-3.5}{4.4}...\frac{-99.101}{100.100}\)

\(A=\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-99\right)}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}\)

\(A=-\frac{99}{100}.\frac{101}{2}=-\frac{9999}{200}\)

Vi \(-\frac{9999}{200}< -\frac{100}{200}=-\frac{1}{2}\)

V...\(A< -\frac{1}{2}\)

Answer 5

1-5= 111141111111

Answer 6

lộn nha 1-5 =4

Answer 7

\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2-1}\right).\left(\frac{1}{4^2-1}\right)...\left(\frac{1}{100^2-1}\right)\)

\(A=-\left[\frac{1.3.2.4....99.101}{2^2.3^2...100^2}\right]\)

\(A=-\left(\frac{1.2.3...99}{2.3...100}\right).\left(\frac{3.4....101}{2.3....100}\right)\)

\(A=-\frac{1}{100}.\frac{101}{2}=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)

Vay...

Answer 8

\(A=(\frac{1}{2^2}-1)(\frac{1}{3^2}-1)...(\frac{1}{100^2})\)

\(A=(\frac{-1.3}{2.2})(\frac{-2.4}{3.3})...(\frac{-99.101}{100.100})\)

\(A={-(1.3.2.4...99.101)\over2.2.3.3...100.100}\)

\(A=-(\frac{1.2...99}{2.3.100}).(\frac{3.4...101}{2.3...100})\)

\(A=\frac{-1}{100}.\frac{101}{2}=\frac{-101}{100}<\frac{-50}{100}=\frac{-1}{2}\)

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