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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=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
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 = \(\dfrac{1}{101}\)
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 \(\dfrac{1}{50}< \dfrac{1}{49}\)
\(\Rightarrow A^2< \dfrac{1}{49}\)
\(\Rightarrow A< \dfrac{1}{7}\left(đpcm\right)\)
\(\dfrac{-5}{9}+1\dfrac{5}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{15}{20}-\dfrac{8}{20}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\dfrac{7}{20}:49\\ =\dfrac{-5}{9}+\dfrac{49}{90}:49\\ =\dfrac{-5}{9}+\dfrac{1}{90}\\ =\dfrac{-50}{90}+\dfrac{1}{90}\\ =\dfrac{-49}{90}\)
\(1\dfrac{13}{15}.0,75-\left(\dfrac{104}{195}+25\%\right).\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{24}{47}-\dfrac{51}{13}:3\\ =\dfrac{7}{5}-\left(\dfrac{32}{60}+\dfrac{15}{60}\right).\dfrac{24}{47}-\dfrac{51}{13}.\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =1-\dfrac{17}{13}\\ =\dfrac{13}{13}-\dfrac{17}{13}\\ =\dfrac{-4}{13}\)
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.............
A,
\(\left(7\dfrac{4}{9}+3\dfrac{7}{11}\right)-3\dfrac{4}{9}=7\dfrac{4}{9}+3\dfrac{7}{11}-3\dfrac{4}{9}\)
\(=7\dfrac{4}{9}-3\dfrac{4}{9}+3\dfrac{7}{11}=4+3\dfrac{7}{11}=7\dfrac{7}{11}\)
B,
\(5\dfrac{2}{7}.\dfrac{8}{11}+5\dfrac{2}{7}.\dfrac{5}{11}-5\dfrac{2}{7}.\dfrac{2}{11}=5\dfrac{2}{7}.\left(\dfrac{8}{11}+\dfrac{5}{11}-\dfrac{2}{11}\right)\)
\(=5\dfrac{2}{7}.1=5\dfrac{2}{7}\)
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: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
\(b)\left(x-3\right)^3=125^2\)
\(\Rightarrow\left(x-3\right)^3=5^{3^2}\)
\(\Rightarrow\left(x-3\right)^3=25^3\)
\(\Rightarrow x-3=25\)
\(\Rightarrow x=28\)
\(a.\)
\(-\dfrac{2}{3}\cdot\dfrac{?}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{?}{4}=\dfrac{1}{2}:-\dfrac{2}{3}=\dfrac{1}{2}\cdot-\dfrac{3}{2}=-\dfrac{3}{4}\)
\(\Leftrightarrow?=-3\)
\(b.\)
\(\dfrac{?}{3}\cdot\dfrac{5}{8}=-\dfrac{5}{12}\)
\(\Leftrightarrow\dfrac{?}{3}=\dfrac{-5}{12}:\dfrac{5}{8}=\dfrac{-5}{12}\cdot\dfrac{8}{5}=-\dfrac{2}{3}\)
\(\Leftrightarrow?=-2\)
\(c.\)
\(\dfrac{5}{6}\cdot\dfrac{3}{?}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{?}=\dfrac{1}{4}:\dfrac{5}{6}=\dfrac{1}{4}\cdot\dfrac{6}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow?=10\)
Mk gọi ? = x nha
a) \(\dfrac{-2}{3}.\dfrac{x}{4}=\dfrac{1}{2}\)
\(\dfrac{x}{4}=\dfrac{1}{2}:\dfrac{-2}{3}\)
\(\dfrac{x}{4}=\dfrac{-3}{4}\)
⇒x=-3
b)\(\dfrac{x}{3}.\dfrac{5}{8}=\dfrac{-5}{12}\)
\(\dfrac{x}{3}=\dfrac{-5}{12}:\dfrac{5}{8}\)
\(\dfrac{x}{3}=\dfrac{-2}{3}\)
⇒x=-2
c)\(\dfrac{5}{6}.\dfrac{3}{x}=\dfrac{1}{4}\)
\(\dfrac{3}{x}=\dfrac{1}{4}:\dfrac{5}{6}\)
\(\dfrac{3}{x}=\dfrac{3}{10}\)
⇒x=10