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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 \(A< -\frac{1}{2}\)
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)...\left(\frac{1}{10000}-1\right)\)
\(=\frac{-3}{4}\cdot\frac{-8}{9}\cdot\frac{-15}{16}\cdot...\cdot\frac{-9999}{10000}\)
\(=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot...\cdot\frac{-99\cdot111}{100.100}\)
\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot...\cdot\frac{99\cdot111}{100\cdot100}\)
\(=\frac{\left(1\cdot2\cdot3\cdot4\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot6\cdot...\cdot111\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot100\right)^2}\)
\(=\frac{101}{2\cdot100}\)
\(=\frac{101}{200}>\frac{1}{2}\)
\(A=-\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).....\left(1-\frac{1}{100^2}\right)\)
\(A=-\left(\frac{1.3}{2.2}\right).\left(\frac{2.4}{3.3}\right)....\left(\frac{99.101}{100.100}\right)\)
\(A=-\left(\frac{1.2....99}{2.3...100}\right).\left(\frac{3.4....101}{2.3....100}\right)\)
\(A=-\left(\frac{1}{100}\right).\left(\frac{101}{2}\right)\)
\(A=\frac{-101}{200}>\frac{-1}{2}\)
A= -(3/4 . 8/9 . 15/16 ... 9999/10000)
=-( (3.8.15....9999)/(4.9.16...10000))
= - (( 1.3.2.4.3.5...99.101)/(2.2.3.3.4.4...100.100))
= -( ( 1.2.3.4...99).(3.4.5..101) / (2.3.4...100) . (2.3.4..100))
= -101/200< -100/200 = -1/2
Vậy A < -1/2
Ta có :
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{100^2}-1\right)\)
\(A< \left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{100}-1\right)\)
\(\Rightarrow A< \left(\frac{-1}{2}\right).\left(\frac{-2}{3}\right)....\left(\frac{-99}{100}\right)\)
\(\Rightarrow A< -\left(\frac{1}{2}.\frac{2}{3}...\frac{99}{100}\right)\)
\(A< -\left(\frac{1.2....99}{2.3...100}\right)=\frac{-1}{100}\)
\(\)Mà \(\frac{-1}{100}>\frac{-1}{2}\)
\(\Rightarrow A>\frac{-1}{2}\)
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\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(A=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-9999}{100^2}\)
\(A=-\left(\frac{3}{2^2}.\frac{8}{3^2}...\frac{9999}{100^2}\right)\)(vì có 99 thừa số, mỗi thừa số là âm nên kết quả là âm)
\(A=-\left(\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{99.101}{100.100}\right)\)
\(A=-\left(\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}\right)\)
\(A=-\left(\frac{1}{100}.\frac{101}{2}\right)=-\frac{101}{200}\)
\(A< -\frac{100}{200}=-\frac{1}{2}\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{100^2-1}{100^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{99.101}{100.100}\)
\(=\frac{\left(1.2.3...99\right)\left(2.3.4...101\right)}{\left(2.3.4...100\right)\left(2.3.4...100\right)}\)
\(=\frac{101}{100}\)
mình lam hơi sai mà cũng khá lac đề, bạn từ bài của mình mà làm bài khác đúng hơn nha hiền
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
câu hỏi của Lê Vũ Anh Thư nhé, vừa đc đăng lên
chúc hok tốt!
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(\Rightarrow A=\left(\frac{1}{2^2}-\frac{4}{2^2}\right)\left(\frac{1}{3^2}-\frac{9}{3^2}\right)...\left(\frac{1}{100^2}-\frac{10000}{100^2}\right)\)
\(\Rightarrow A=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-9999}{100^2}\)
\(\Rightarrow A=-\frac{3}{2^2}.\frac{8}{3^2}...\frac{9999}{100^2}\)
\(\Rightarrow A=-\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{99.101}{100.100}\)
\(\Rightarrow A=-\frac{\left(1.2...99\right)\left(3.4...101\right)}{\left(2.3...100\right)\left(2.3...100\right)}\)
\(\Rightarrow A=-\frac{101}{100.2}=\frac{-101}{200}< \frac{-100}{200}=\frac{-1}{2}\)
Vậy \(A< \frac{-1}{2}\)