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\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)
\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)
\(A=\left[35-0\right]-5\frac{7}{32}\)
\(A=35-5\frac{7}{32}\)
\(A=\frac{953}{32}\)
\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)
\(B=71\frac{38}{45}-\frac{36377}{855}\)
\(B=\frac{1670}{57}\)
\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\frac{153}{14}:\frac{4}{5}\)
\(C=\frac{765}{56}\)
\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)
\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0-\frac{1}{4}\)
\(D=-\frac{1}{4}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)
\(\)\(E=\frac{22}{45}\)
CHUC BAN HOC TOT >.<
a) \(\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{23}\right)=\frac{31}{23}-\frac{7}{32}-\frac{8}{23}=1-\frac{7}{32}=\frac{25}{32}\)
b) \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}-\left(\frac{79}{67}-\frac{12}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(=\frac{1}{3}-1+1=\frac{1}{3}\)
d) \(\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{-1}{3}+\frac{17}{19}=\frac{1}{7}.\left(\frac{1}{3}-\frac{1}{3}\right)+\frac{17}{19}=\frac{17}{19}\)
e) \(\frac{3}{5}.\frac{7}{9}+\frac{7}{5}.\frac{2}{9}=\frac{7}{5}.\left(\frac{3}{9}+\frac{2}{9}\right)=\frac{7}{5}.\frac{5}{9}=\frac{7}{9}\)
a) \(\frac{1}{3}.\frac{-6}{13}.\frac{-9}{10}.\frac{-13}{36}\)
\(=\left(\frac{1}{3}.\frac{-9}{10}\right)\left(\frac{-6}{13}.\frac{-13}{36}\right)\)
\(=\frac{-3}{10}.\frac{1}{6}\)
\(=\frac{-1}{20}\)
b) \(\frac{-1}{3}.\frac{-15}{17}.\frac{34}{45}\)
\(=\frac{-1}{3}.\frac{-2}{3}\)
\(=\frac{2}{9}\)
c) \(\left(1-\frac{1}{5}\right)\left(\frac{-3}{10}+\frac{1}{5}\right)\)
\(=\frac{4}{5}.\frac{-1}{10}\)
\(=\frac{-2}{25}\)
d) \(A=\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}+\frac{2}{3}\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)+\frac{2}{3}\)
\(=\frac{1}{3}.2+\frac{2}{3}\)
\(=\frac{2}{3}+\frac{2}{3}\)
\(=\frac{4}{3}\)
e) \(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)
\(=\left(11\frac{1}{4}-5\frac{1}{4}\right)-2\frac{5}{7}\)
\(=6-2\frac{5}{7}\)
\(=5\frac{7}{7}-2\frac{5}{7}\)
\(=3\frac{2}{7}\)
\(\frac{37}{45}-\left(\frac{7}{45}-\frac{36}{108}-\frac{5}{13}\right)=\frac{37}{45}-\frac{7}{45}+\frac{36}{108}+\frac{5}{13}.\)
\(=\left(\frac{37}{45}-\frac{7}{45}\right)+\left(\frac{36}{108}+\frac{5}{13}\right)\)
\(=\frac{2}{3}+\frac{28}{39}\)
\(=\frac{18}{13}\)
\((\frac{37}{45}-\frac{7}{45})+(\frac{36}{108}-\frac{5}{13})\)
=\(\frac{2}{3}+-\frac{2}{39}\)=\(\frac{8}{13}\)