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b) Vì \(\left|x+\dfrac{1}{1.3}\right| \ge0;\left|x+\dfrac{1}{3.5}\right|\ge0;...;\left|x+\dfrac{1}{97.99}\right|\ge0\)
\(\Rightarrow50x\ge0\Rightarrow x\ge0\)
Khi đó: \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3};\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5};...;\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\left(1\right)\)
Thay (1) vào đề bài:
\(x+\dfrac{1}{1.3}+x+\dfrac{1}{3.5}+...+x+\dfrac{1}{97.99}=50x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)=50x\)
\(\Rightarrow49x+\left[\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\right]=50x\)
\(\Rightarrow49x+\dfrac{16}{99}=50x\)
\(\Rightarrow x=\dfrac{16}{99}\)
Vậy \(x=\dfrac{16}{99}.\)
Giải:
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
Đk: \(n\ne0;n\ne-1\)
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\left(\dfrac{2.3-2}{2.3}\right)\left(\dfrac{3.4-2}{3.4}\right)\left(\dfrac{4.5-2}{4.5}\right)...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{4}{2.3}.\dfrac{10}{3.4}.\dfrac{18}{4.5}...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\left(\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4.2.5.3.6...\left(n-1\right)\left(n+2\right)}{2.3.3.4.4.5.n\left(n+1\right)}\)
\(\Leftrightarrow C=\dfrac{\left[1.2.3...\left(n-1\right)\right]\left[4.5.6\left(n+2\right)\right]}{\left(2.3.4...n\right)\left[3.4.5....\left(n+1\right)\right]}\)
\(\Leftrightarrow C=\dfrac{n+2}{3n}\)
Vì \(\dfrac{n+2}{3n}< \dfrac{2n+2}{3n}\)
\(\Leftrightarrow C< \dfrac{2n+2}{3n}\)
Vậy ...
Giải:
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
Đk: \(n\ne0;n\ne-1\)
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\left(\dfrac{2.3-2}{2.3}\right)\left(\dfrac{3.4-2}{3.4}\right)\left(\dfrac{4.5-2}{4.5}\right)...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{4}{2.3}.\dfrac{10}{3.4}.\dfrac{18}{4.5}...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\left(\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4.2.5.3.6...\left(n-1\right)\left(n+2\right)}{2.3.3.4.4.5.n\left(n+1\right)}\)
\(\Leftrightarrow C=\dfrac{\left[1.2.3...\left(n-1\right)\right]\left[4.5.6\left(n+2\right)\right]}{\left(2.3.4...n\right)\left[3.4.5....\left(n+1\right)\right]}\)
\(\Leftrightarrow C=\dfrac{n+2}{3n}\)
Vì \(\dfrac{n+2}{3n}< \dfrac{2n+2}{3n}\)
\(\Leftrightarrow C< \dfrac{2n+2}{3n}\)
Vậy ...
Sửa đề:
\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)....\left(\dfrac{1}{100^2}-1\right)\)
\(A=\left(\dfrac{1}{2^2}-\dfrac{2^2}{2^2}\right)\left(\dfrac{1}{3^2}-\dfrac{3^2}{3^2}\right)\left(\dfrac{1}{4^2}-\dfrac{4^2}{4^2}\right)....\left(\dfrac{1}{100^2}-\dfrac{100^2}{100^2}\right)\)
\(A=\dfrac{\left(1-2^2\right)}{2^2}.\dfrac{\left(1-3^2\right)}{3^2}.\dfrac{\left(1-4^2\right)}{4^2}....\dfrac{\left(1-100^2\right)}{100^2}\)
\(A=\dfrac{\left(1-2\right)\left(1+2\right)}{2^2}.\dfrac{\left(1-3\right)\left(1+3\right)}{3^2}.\dfrac{\left(1-4\right)\left(1+4\right)}{4^2}......\dfrac{\left(1-100\right)\left(1+100\right)}{100^2}\)
\(A=\dfrac{-3}{2^2}.\dfrac{-8}{3^2}.\dfrac{-15}{4^2}....\dfrac{-9999}{100^2}\)
Ta xét từ \(2\) đến \(100\) có: \(\dfrac{\left(100-2\right)}{1}+1=99\)
\(50\) là số lẻ nên tích trên là số âm
Hay \(-A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{99.101}{100.100}\)
\(-A=\dfrac{1.3.2.4.3.5....99.101}{2.2.3.3.4.4.....100.100}\)
\(-A=\dfrac{1.2.3....99}{2.3.4....100}.\dfrac{3.4.5....101}{2.3.4....100}\)
\(-A=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}\)
\(A=-\dfrac{101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
a: \(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\)
\(=-\dfrac{1}{10}\)
9<10
=>1/9>1/10
=>\(-\dfrac{1}{9}< -\dfrac{1}{10}\)
=>\(A>-\dfrac{1}{9}\)
b: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{10}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{11}{10}\)
\(=\dfrac{-1}{10}\cdot\dfrac{11}{2}=\dfrac{-11}{20}\)
20<21
=>\(\dfrac{11}{20}>\dfrac{11}{21}\)
=>\(-\dfrac{11}{20}< -\dfrac{11}{21}\)
=>\(B< -\dfrac{11}{21}\)
\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)..............\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right).............\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}.\dfrac{3}{2}.\dfrac{-2}{3}.\dfrac{4}{3}.............\dfrac{-99}{100}.\dfrac{101}{100}\)
\(=\dfrac{-\left(1.2.3....99\right)}{2.3......100}.\dfrac{3.4...101}{2.3....100}\)
\(=\dfrac{-1}{100}.\dfrac{101}{2}\)
\(=\dfrac{-101}{200}< \dfrac{-1}{2}\)
\(\Leftrightarrow A< \dfrac{-1}{2}\)
\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)...\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}.\dfrac{3}{2}.\dfrac{-2}{3}.\dfrac{4}{3}...\dfrac{-99}{100}.\dfrac{101}{100}\)
\(=\dfrac{-\left(1.2...99\right)}{2.3...100}.\dfrac{3.4...101}{2.3...100}=\dfrac{-1}{100}.\dfrac{101}{2}\)
\(=\dfrac{-101}{200}< \dfrac{-1}{2}\)
\(\Rightarrow A< \dfrac{-1}{2}\)
\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right).....\left(\dfrac{1}{100^2}-1\right)\)
\(A=-\dfrac{3}{4}.-\dfrac{8}{9}.-\dfrac{15}{16}.....-\dfrac{9999}{10000}\)
\(A=\dfrac{-\left(1.3\right)}{2.2}.\dfrac{-\left(2.4\right)}{3.3}.\dfrac{-\left(3.5\right)}{4.4}......\dfrac{-\left(99.101\right)}{100.100}\)
Từ \(-1\) đến \(-99\) có: \(\left(99-1\right):1+1=99\)
Cộng thêm số \(-101\) tất cả có \(100\) số hạng
\(A=\dfrac{1.3.2.4.3.5....99.101}{2.2.3.3.4.4.....100.100}\)
\(A=\dfrac{1.2.3....99}{2.3.4....100}.\dfrac{3.4.5....101}{2.3.4....100}\)
\(A=\dfrac{1}{100}.\dfrac{101}{2}\)
\(A=\dfrac{101}{200}\)
\(\)
A = \(\dfrac{-101}{200}>\dfrac{-100}{200}=\dfrac{-1}{2}\)
Nhận thấy A có 99 hạng tử mà mỗi hạng tử chứa dấu âm nên viết gọn\(A=-\dfrac{3}{4}.\dfrac{8}{9}.....\dfrac{9999}{10000}=-\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}....\dfrac{99.101}{100^2}=-\dfrac{\left(1.2...99\right).\left(3.4...101\right)}{\left(2.3..100\right).\left(2.3...100\right)}=-\dfrac{101}{2.100}=-\dfrac{101}{200}< -\dfrac{1}{2}\)