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a ) \(\frac{2}{5}+\frac{-1}{6}-\frac{3}{4}-\frac{-2}{3}\)
= \(\frac{2}{5}+\frac{-1}{6}+\frac{-3}{4}+\frac{2}{3}\)
= \(\left(\frac{2}{5}+\frac{-3}{4}\right)+\left(\frac{-1}{6}+\frac{2}{3}\right)\)
= \(\left(\frac{8}{20}+\frac{-15}{20}\right)+\left(\frac{-1}{6}+\frac{4}{6}\right)\)
= \(\left(\frac{8+\left(-15\right)}{20}\right)+\left(\frac{\left(-1\right)+4}{6}\right)\)
= \(\frac{-7}{20}+\frac{1}{2}\)
= \(\frac{-7}{20}+\frac{10}{20}=\frac{\left(7\right)+10}{20}=\frac{3}{20}\)
tk mk nha
đang âm rất nhiều rồi , giúp nha !!!!!
a) Ta có: \(\frac{-1}{12}-\left(2\frac{5}{8}-\frac{1}{3}\right)\)
\(=-\frac{1}{12}-\frac{21}{8}+\frac{1}{3}\)
\(=\frac{-6}{72}-\frac{189}{72}+\frac{24}{72}\)
\(=-\frac{19}{8}\)
b) Ta có: \(-1,75-\left(\frac{-1}{9}-2\frac{1}{18}\right)\)
\(=\frac{-7}{4}+\frac{1}{9}+\frac{37}{18}\)
\(=\frac{-63}{36}+\frac{4}{36}+\frac{74}{36}\)
\(=\frac{5}{12}\)
c) Ta có: \(\frac{2}{5}+\frac{-4}{3}+\frac{-1}{2}\)
\(=\frac{12}{30}+\frac{-40}{30}+\frac{-15}{30}\)
\(=-\frac{43}{30}\)
d) Ta có: \(\frac{3}{12}-\left(\frac{6}{15}-\frac{3}{10}\right)\)
\(=\frac{3}{12}-\frac{6}{15}+\frac{3}{10}\)
\(=\frac{15}{60}-\frac{24}{60}+\frac{18}{60}\)
\(=\frac{3}{20}\)
e) Ta có: \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)
\(=\frac{93}{11}+\frac{29}{8}-\frac{38}{11}\)
\(=5+\frac{29}{8}=\frac{40}{8}+\frac{29}{8}=\frac{69}{8}\)
f) Ta có: \(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
\(=\frac{4}{9}\cdot\left(-7\right)+\frac{59}{9}\cdot\left(-7\right)\)
\(=\left(-7\right)\cdot\left(\frac{4}{9}+\frac{59}{9}\right)=\left(-7\right)\cdot7=-49\)
g) Ta có: \(\frac{-1}{4}\cdot13\frac{9}{11}-0,25\cdot6\frac{2}{11}\)
\(=\frac{-1}{4}\cdot\frac{152}{11}+\frac{-1}{4}\cdot\frac{68}{11}\)
\(=\frac{-1}{4}\cdot\left(\frac{152}{11}+\frac{68}{11}\right)=-\frac{1}{4}\cdot20=-5\)
h) Ta có: \(5\frac{27}{5}+\frac{27}{23}+0,5-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{52}{5}+\frac{27}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{52}{5}+\frac{43}{23}+\frac{1}{2}-\frac{5}{27}\)
\(=\frac{64584}{6210}+\frac{11610}{6210}+\frac{3105}{6210}-\frac{1150}{6210}\)
\(=\frac{78149}{6210}\)
i) Ta có: \(\frac{3}{8}\cdot27\frac{1}{5}-51\frac{1}{5}\cdot\frac{3}{8}+19\)
\(=\frac{3}{8}\cdot\frac{136}{5}-\frac{3}{8}\cdot\frac{206}{5}+\frac{3}{8}\cdot\frac{152}{3}\)
\(=\frac{3}{8}\cdot\left(\frac{136}{5}-\frac{206}{5}+\frac{152}{3}\right)=\frac{3}{8}\cdot\frac{110}{3}\)
\(=\frac{55}{4}\)
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) \(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)
\(=\left(11\frac{1}{4}-5\frac{1}{4}\right)-2\frac{5}{7}\)
\(=6-2\frac{5}{7}\)
\(=5\frac{7}{7}-2\frac{5}{7}\)
\(=3\frac{2}{7}\)