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\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\\ 2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\\ 2A-A=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\\ A=1-\dfrac{1}{2^9}=\dfrac{511}{512}\)
\(B=\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\\ 3B=\dfrac{3}{4}+\dfrac{3}{12}+\dfrac{3}{36}+\dfrac{3}{108}+\dfrac{3}{324}+\dfrac{3}{972}\\ 3B=\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\\ 3B-B=\left(\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\right)-\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\right)\\ 2B=\dfrac{3}{4}-\dfrac{1}{972}=\dfrac{182}{243}\\ B=\dfrac{364}{243}\)
a) \(\dfrac{1+\dfrac{1}{4}}{1-\dfrac{1}{4}}:\dfrac{1+\dfrac{1}{8}}{1-\dfrac{1}{8}}\\ =\dfrac{\dfrac{5}{4}}{\dfrac{3}{4}}:\dfrac{\dfrac{9}{8}}{\dfrac{7}{8}}\\ =\dfrac{5}{3}:\dfrac{9}{7}\\ =\dfrac{5}{3}.\dfrac{9}{7}\\ =\dfrac{35}{27}=1\dfrac{8}{27}\)
b) \(0,25+37\%-2\dfrac{1}{4}\\ =\dfrac{1}{4}+\dfrac{37}{100}-\dfrac{9}{4}\\ =\dfrac{25+37-225}{100}\\ =-\dfrac{163}{100}=-1\dfrac{63}{100}\)
a, A= 1/2. (2/1.2.3+2/2.3.4+2/3.4.5+...+2/18.19.20) A=1/2. (1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/18.19-1/19.20) A=1/2. (1/1.2-1/19.20) A=1/2. 189/380 A= 189/760
3) \(\dfrac{2}{3}-\left(-\dfrac{1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(-\dfrac{5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{90}\)
= \(\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{90}\)
= \(\left(\dfrac{2}{3}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{90}\right)+\left(\dfrac{1}{4}+\dfrac{1}{12}\right)\)
= \(\left(\dfrac{60}{90}+\dfrac{54}{90}-\dfrac{14}{90}+\dfrac{50}{90}+\dfrac{1}{9}\right)+\left(\dfrac{4}{12}+\dfrac{1}{12}\right)\)
= \(\dfrac{151}{90}+\dfrac{1}{3}=\dfrac{151}{90}+\dfrac{30}{90}=\dfrac{181}{90}\)
a) \(\dfrac{1}{3}+\dfrac{3}{8}-\dfrac{7}{12}\)
\(=\dfrac{17}{24}-\dfrac{7}{12}\)
\(=\dfrac{1}{8}\)
b) \(\dfrac{-3}{14}+\dfrac{5}{8}-\dfrac{1}{2}\)
\(=\dfrac{23}{56}-\dfrac{1}{2}\)
\(=\dfrac{-5}{56}\)
c) \(\dfrac{1}{4}-\dfrac{2}{3}-\dfrac{11}{18}\)
\(=\dfrac{-5}{12}-\dfrac{11}{18}\)
\(=\dfrac{-37}{36}\)
d) \(\dfrac{1}{4}+\dfrac{5}{12}-\dfrac{1}{13}-\dfrac{7}{8}\)
\(=\dfrac{2}{3}-\dfrac{1}{13}-\dfrac{7}{8}\)
\(=\dfrac{23}{39}-\dfrac{7}{8}\)
\(=\dfrac{-89}{312}\)
a) \(1-\dfrac{1}{2}=\dfrac{1}{2}\)
\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3-2}{6}=\dfrac{1}{6}\)
\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4-3}{12}=\dfrac{1}{12}\)
\(\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{5-4}{20}=\dfrac{1}{20}\)
\(\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{6-5}{30}=\dfrac{1}{30}\)
b) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}\right)\)
\(=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+-\dfrac{1}{6}\)\(=1+-\dfrac{1}{6}\)
\(=\dfrac{5}{6}\)
\(B=\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\\\)
\(3B=3\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\right)\)
\(3B=\dfrac{3}{4}+\dfrac{3}{12}+\dfrac{3}{36}+\dfrac{3}{108}+\dfrac{3}{324}+\dfrac{3}{972}\)
\(3B=\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\)
\(2B=3B-B\)
\(2B=\left(\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\right)-\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\right)\)
\(2B=\dfrac{3}{4}-\dfrac{1}{972}=\dfrac{729-1}{972}=\dfrac{728}{972}=\dfrac{182}{243}\)
\(B=\dfrac{182}{243}:\dfrac{1}{2}=\dfrac{182\cdot2}{243}=\dfrac{364}{243}\)