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Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)
a) \(\left(1\dfrac{1}{2}\right)\left(1\dfrac{1}{3}\right)..............\left(1\dfrac{1}{100}\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}....................\dfrac{101}{100}\)
\(=\dfrac{1}{2}.\dfrac{101}{1}=\dfrac{101}{2}\)
b) \(1\dfrac{1}{2}.1\dfrac{1}{3}.1\dfrac{1}{4}...................1\dfrac{1}{2007}\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}....................\dfrac{2008}{2007}\)
\(=\dfrac{1}{2}.\dfrac{2008}{1}=1004\)
c) \(1\dfrac{1}{2}.1\dfrac{1}{3}.....................1\dfrac{1}{2017}\)
\(=\dfrac{3}{2}.\dfrac{4}{3}..................\dfrac{2018}{2017}\)
\(=\dfrac{1}{2}.\dfrac{2018}{1}=1009\)
a,
\(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)+\sqrt{2}\cdot\dfrac{\sqrt{2^5}}{1-\sqrt{9}}\)
\(=2^2-\left(\sqrt{3}\right)^2+\dfrac{\sqrt{2}\cdot\sqrt{2^5}}{1-3}=4-3+\dfrac{\sqrt{2^6}}{-2}=1+\dfrac{8}{-2}=1+\left(-4\right)=-3\)
b,
\(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}\right)\cdot\dfrac{49}{50}\)
\(=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\cdot\dfrac{49}{50}\)
\(=\left(1-\dfrac{1}{50}\right)\cdot\dfrac{49}{50}=\dfrac{49}{50}\cdot\dfrac{49}{50}=\dfrac{49^2}{50^2}=\dfrac{2401}{2500}\)
(31/5:217/10-1,2*125)*7/4-1/2
=(2/7-150)*7/4-1/2
=-1048/7*7/4-1/2
=-262-1/2
=-525/2
\(\left(6\dfrac{1}{5}:21,7-1,2\cdot5^3\right)\cdot1\dfrac{3}{4}-\dfrac{1}{2}\)
\(=\left(\dfrac{31}{5}\cdot\dfrac{10}{217}-1,2\cdot125\right)\cdot1,75-0,5\)
\(=\left(\dfrac{62}{217}-150\right)\cdot1,75-0,5\)
\(=\dfrac{62}{217}\cdot1,75-150\cdot1,75-0,5\)
\(=0,5-262,5-0,5\)
\(=-262,5\)
\(a)\dfrac{-5}{21}-\dfrac{1}{3}+3\dfrac{1}{2}.\left(\dfrac{-2}{3}\right)^3\)
\(=\dfrac{-5}{21}+\dfrac{-7}{21}+\dfrac{7}{2}.\dfrac{-8}{27}\)
\(=-\dfrac{4}{7}+\dfrac{-28}{27}\)
\(=\dfrac{-108}{189}+\dfrac{-196}{189}\)
\(=-\dfrac{304}{189}\)
\(b)-2\dfrac{1}{3}+\left(\dfrac{3}{8}-\dfrac{3}{4}\right)^3:\dfrac{5}{9}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\left(\dfrac{3}{8}-\dfrac{6}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\left(-\dfrac{3}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\dfrac{-27}{512}.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\dfrac{-243}{2560}-\dfrac{1}{2}\)
\(=\dfrac{-17920}{7680}+\dfrac{-729}{7680}+\dfrac{-3840}{7680}\)
\(=\dfrac{-22489}{7680}\)
\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)....................\left(\dfrac{1}{10^2}-1\right)\)
\(=\left(\dfrac{1}{4}-\dfrac{4}{4}\right)\left(\dfrac{1}{9}-\dfrac{9}{9}\right)...........\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)
\(=\dfrac{-3}{4}.\dfrac{-8}{9}..............\dfrac{-99}{100}\)
\(=\dfrac{\left(-1\right).3}{2.2}.\dfrac{\left(-2\right).4}{3.3}..................\dfrac{\left(-9\right).11}{10.10}\)
\(=\dfrac{\left(-1\right)\left(-2\right)..........\left(-9\right)}{2.3.....10}.\dfrac{3.4....11}{2.3....10}\)
\(=\dfrac{-1}{10}.\dfrac{11}{2}\)
\(=\dfrac{-11}{20}< \dfrac{-10}{20}=\dfrac{-1}{2}\)
\(\Leftrightarrow A< \dfrac{-1}{2}\)
\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)...\left(\dfrac{1}{10^2}-1\right)\)
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right).....\left(\dfrac{1}{100}-1\right)\)
\(A=\left(\dfrac{1}{4}-\dfrac{4}{4}\right)\left(\dfrac{1}{9}-\dfrac{9}{9}\right)....\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)
\(A=\dfrac{-3}{4}.\dfrac{-8}{9}.....\dfrac{-99}{100}\)
\(A=\dfrac{\left(-1\right).3}{4}.\dfrac{\left(-1\right).8}{9}......\dfrac{\left(-1\right).99}{100}\)
\(A=\dfrac{\left(-1\right).1.3}{2.2}.\dfrac{-1.2.4}{3.3}....\dfrac{-1.9.11}{10.10}\)
\(A=\dfrac{-1.3}{2.2}.\dfrac{-2.4}{3.3}....\dfrac{-9.11}{10.10}\)
\(A=\dfrac{\left(-1\right)\left(-2\right)....\left(-9\right)}{2.3.....10}.\dfrac{3.4....11}{2.3.....10}\)
\(A=\dfrac{-1}{10}.\dfrac{11}{2}=-\dfrac{11}{20}\)
\(\dfrac{1}{5}+\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{6}{30}+\dfrac{15}{30}-\dfrac{10}{30}\)
\(\dfrac{1}{5}+\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{6}{30}+\dfrac{15}{30}-\dfrac{10}{30}=\dfrac{1}{30}\)