3/5+-7/13*13/21
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\(\dfrac{26}{81}+\dfrac{4}{27}=\dfrac{26+12}{81}=\dfrac{38}{81}\)
\(\dfrac{5}{64}+\dfrac{7}{8}=\dfrac{5+56}{64}=\dfrac{61}{64}\)
\(\dfrac{26}{81}+\dfrac{4}{27}=\dfrac{26}{81}+\dfrac{4\times3}{27\times3}=\dfrac{26}{81}+\dfrac{12}{81}=\dfrac{38}{81}\)
\(\dfrac{5}{64}+\dfrac{7}{8}=\dfrac{5}{64}+\dfrac{7\times8}{8\times8}=\dfrac{5}{64}+\dfrac{56}{64}=\dfrac{61}{64}\)
5.\(-\dfrac{3}{7}+\dfrac{5}{13}+\dfrac{-4}{12}=-\dfrac{103}{273}\)
b.\(-\dfrac{5}{21}+\dfrac{-2}{21}+\dfrac{8}{24}=\dfrac{-5-2}{21}+\dfrac{8}{24}=-\dfrac{7}{21}+\dfrac{8}{24}=-\dfrac{1}{3}+\dfrac{8}{24}=0\)
c.\(\dfrac{5}{13}+\dfrac{-5}{7}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+-\dfrac{5}{7}=1-1-\dfrac{5}{7}=-\dfrac{5}{7}\)
a: =-5/11-6/11+1=-11/11+1=0
b: =-13/17-13/21-4/17=-1-13/21=-34/21
b: \(=-\dfrac{5}{12}\cdot\dfrac{9}{20}\cdot\dfrac{7}{17}=\dfrac{-21}{272}\)
d: \(=\dfrac{13}{17}\left(-\dfrac{4}{5}-\dfrac{3}{4}\right)=\dfrac{13}{17}\cdot\dfrac{-31}{20}=\dfrac{-403}{340}\)
\(\left(\dfrac{7}{3}-\dfrac{5}{21}\right)-\left(\dfrac{16}{21}+\dfrac{20}{13}-\dfrac{1}{5}\right)\)
\(=\left(\dfrac{49}{21}-\dfrac{5}{21}\right)-\left(\dfrac{1040}{1365}+\dfrac{2100}{1365}-\dfrac{273}{1365}\right)\)
\(=\dfrac{44}{21}-\left(\dfrac{2867}{1365}\right)\)
\(=-\dfrac{1}{195}\)
\(a,\dfrac{2}{5}-\dfrac{3}{8}+\dfrac{7}{6}=\dfrac{16}{40}-\dfrac{15}{40}+\dfrac{7}{6}=\dfrac{1}{40}+\dfrac{7}{6}=\dfrac{3}{120}+\dfrac{140}{120}=\dfrac{143}{120}\)
\(b,\dfrac{2}{3}+\left[\dfrac{5}{7}+\left(-\dfrac{2}{3}\right)\right]=\dfrac{2}{3}+\dfrac{5}{7}+\left(-\dfrac{2}{3}\right)=\left[\dfrac{2}{3}+\left(-\dfrac{2}{3}\right)\right]+\dfrac{5}{7}=0+\dfrac{5}{7}=\dfrac{5}{7}\)
\(c,\dfrac{5}{13}+\left(-\dfrac{5}{7}\right)+\left(-\dfrac{20}{41}\right)+\dfrac{8}{13}+\left(-\dfrac{21}{41}\right)\)
\(=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)-\dfrac{5}{7}-\dfrac{20}{41}-\dfrac{21}{41}=-\dfrac{5}{7}\)
a: =48/120-45/120+140/120
=143/120
b: =2/3-2/3+5/7=5/7
c: =5/13+8/13-5/7-20/41-21/41
=-5/7
\(\dfrac{3}{5}+\dfrac{-7}{13}\cdot\dfrac{13}{21}\)
\(=\dfrac{3}{5}-\dfrac{7}{21}\)
\(=\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{9-5}{15}=\dfrac{4}{15}\)