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a) \(\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10}{15}+\dfrac{9}{15}=\dfrac{19}{15}\)
a) \(\dfrac{7}{12}-\dfrac{2}{7}+\dfrac{1}{12}=\dfrac{2}{3}-\dfrac{2}{7}=\dfrac{14}{21}-\dfrac{6}{21}=\dfrac{8}{21}\)
\(\dfrac{2}{5}+\dfrac{4}{9}=\dfrac{18}{45}+\dfrac{20}{45}=\dfrac{18+20}{45}=\dfrac{38}{45}\)
a: Ta có: \(A=\frac{-2}{9}+\frac{-3}{4}+\frac35+\frac{1}{15}+\frac{1}{57}+\frac13+\frac{-1}{36}\)
\(=\left(-\frac29-\frac34-\frac{1}{36}\right)+\left(\frac35+\frac{1}{15}+\frac13\right)+\frac{1}{57}\)
\(=\left(-\frac{8}{36}-\frac{27}{36}-\frac{1}{36}\right)+\left(\frac{9}{15}+\frac{1}{15}+\frac{5}{15}\right)+\frac{1}{57}\)
\(=-1+1+\frac{1}{57}=\frac{1}{57}\)
b: \(B=\frac12+\frac{-1}{5}+\frac{-5}{7}+\frac16+\frac{-3}{35}+\frac13+\frac{1}{41}\)
\(=\left(\frac12+\frac13+\frac16\right)-\left(\frac15+\frac57+\frac{3}{35}\right)+\frac{1}{41}\)
\(=\left(\frac36+\frac26+\frac16\right)-\left(\frac{7}{35}+\frac{25}{35}+\frac{3}{35}\right)+\frac{1}{41}\)
\(=1-1+\frac{1}{41}=\frac{1}{41}\)
c: \(C=\frac{-1}{2}+\frac35+\frac{-1}{9}+\frac{1}{127}+\frac{-7}{18}+\frac{4}{35}+\frac27\)
\(=\left(-\frac12-\frac19-\frac{7}{18}\right)+\left(\frac35+\frac27+\frac{4}{35}\right)+\frac{1}{127}\)
\(=\left(-\frac{9}{18}-\frac{2}{18}-\frac{7}{18}\right)+\left(\frac{21}{35}+\frac{10}{35}+\frac{4}{35}\right)+\frac{1}{127}\)
\(=-1+1+\frac{1}{127}=\frac{1}{127}\)
TL:
\(\frac{12}{100}\)= 0,12
\(\frac{5}{100}\)= 0,05
\(\frac{306}{1000}\)= 0,306
-HT-
a; A = \(\dfrac{4026\times2014+4030}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2014+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2013\times2+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-4026+2015\right)}{2013\times2016-2011}\)
A = \(\dfrac{2\times\left(2013\times2016-2011\right)}{2013\times2016-2011}\)
A = 2
\(B=\dfrac{5}{3}x\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{90}\right)=\)
\(=\dfrac{5}{3}x\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...+\dfrac{1}{9x10}\right)\)
\(C=\dfrac{2-1}{1x2}+\dfrac{3-2}{2x3}+\dfrac{4-3}{3x4}+...+\dfrac{10-9}{9x10}=\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}=\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
\(\Rightarrow B=\dfrac{5}{3}xC=\dfrac{5}{3}x\dfrac{9}{10}=\dfrac{3}{2}\)