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Ta có : M . N = \(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{100}{101}\)
= \(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{100}{101}\)
= \(\dfrac{1}{101}\)
Vậy M . N = \(\dfrac{1}{101}\)
a)
\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
Ta có:\(A=\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{98}{99}\)
\(A< \dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\dfrac{7}{8}\cdot...\cdot\dfrac{99}{100}\)
\(\Rightarrow A^2< \dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}\cdot\dfrac{7}{8}\cdot...\cdot\dfrac{98}{99}\cdot\dfrac{99}{100}\)
\(A^2< \dfrac{2}{100}=\dfrac{1}{50}\)
Mà \(\dfrac{1}{50}< \dfrac{1}{49}\)
\(\Rightarrow A^2< \dfrac{1}{49}\)
\(\Rightarrow A< \dfrac{1}{7}\left(đpcm\right)\)
Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
\(\dfrac{6:\dfrac{3}{5}-1\dfrac{1}{6}\cdot\dfrac{6}{7}}{4\dfrac{1}{5}\cdot\dfrac{10}{11}+5\dfrac{2}{11}}\)
\(=\dfrac{6\cdot\dfrac{5}{3}-\dfrac{7}{6}\cdot\dfrac{6}{7}}{\dfrac{21}{5}\cdot\dfrac{10}{11}+\dfrac{57}{11}}\)
\(=\dfrac{10-1}{462+\dfrac{57}{11}}\)
\(=\dfrac{9}{467,\left(18\right)}\)
Thao quy ước của 1 phân số lớn hơn 0 thì:
\(\dfrac{a}{b}>0=>\dfrac{a}{b}< \dfrac{a+n}{b+n}\left(n\ne0\right)\)
Áp dụng vào từng phân số trên ta có: ( các phân số trên lớn hơn 0 nên):
để ý rằng các phân số trên đều lớn hơn 1/100
=>tích cũng lớn hơn 1/100
=>A>1/100
CHÚC BẠN HỌC TỐT.............