\(\frac{-4}{13}.\frac{5}{17}+\frac{-12}{13}.\frac{4}{17}+\frac{4}{13}\)
K
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23 tháng 11 2019
\(\frac{-4}{13}.\frac{5}{17}+\frac{-12}{13}.\frac{4}{17}+\frac{4}{13}\)
\(=\frac{4}{13}.\frac{-5}{17}+\frac{-12}{17}.\frac{4}{13}+\frac{4}{13}\)
\(=\frac{4}{13}.\left(\frac{-5}{17}+\frac{-12}{17}+1\right)\)
\(=\frac{4}{13}.0\)
\(=0\)
31 tháng 8 2017
\(\left(\frac{-4}{13}+\frac{4}{13}+\frac{-12}{13}\right)\times\left(\frac{5}{17}+\frac{4}{17}\right)\)
\(=\)\(\frac{-12}{13}\times\left(\frac{5}{17}+\frac{4}{17}\right)\)
\(=\)\(\frac{-12}{13}\times\frac{9}{17}=\frac{-106}{221}\)
KN
28 tháng 3 2019
\(B=\frac{3+\frac{8}{12}+\frac{9}{13}-\frac{12}{17}}{4+\frac{8}{12}+\frac{12}{13}-\frac{16}{17}}+\frac{4+\frac{16}{60}-\frac{24}{213}-\frac{32}{11}}{5+\frac{20}{61}-\frac{36}{213}-\frac{40}{11}}\)
\(\Leftrightarrow B=\frac{3\left(1+\frac{8}{12}+\frac{3}{13}-\frac{4}{17}\right)}{4\left(1+\frac{8}{12}+\frac{3}{13}-\frac{4}{17}\right)}+\frac{4\left(1+\frac{4}{60}-\frac{6}{213}-\frac{8}{11}\right)}{5\left(1+\frac{4}{60}+\frac{6}{213}-\frac{8}{11}\right)}\)
\(\Leftrightarrow B=\frac{3}{4}+\frac{4}{5}\)
\(\Leftrightarrow B=\frac{15}{20}+\frac{16}{20}\)
\(\Leftrightarrow B=\frac{31}{20}\)
PY
28 tháng 4 2018
a)\(\frac{-10}{13}+\frac{8}{17}-\frac{3}{13}+\frac{12}{17}-\frac{11}{20}\)
= \(\frac{-10}{13}+\frac{8}{17}+\frac{-3}{13}+\frac{12}{17}+\frac{-11}{20}\)
=\(\left(\frac{-10}{13}+\frac{-3}{13}\right)+\left(\frac{8}{17}+\frac{12}{17}\right)+\frac{-11}{20}\)
=\(\frac{-13}{13}+\frac{20}{17}+\frac{-11}{20}\)
= \(\frac{-127}{340}\)
b) \(\frac{3}{4}+\frac{-5}{6}-\frac{11}{-12}\)
= \(\frac{3}{4}+\frac{-5}{6}+\frac{11}{12}\)
= \(\frac{9}{12}+\frac{-10}{12}+\frac{11}{12}\)
=\(\frac{10}{12}=\frac{5}{6}\)
c) \(\left[13.\frac{4}{9}+2.\frac{1}{9}\right]-3.\frac{4}{9}\)
= \(13+2.\left(\frac{4}{9}+\frac{1}{9}\right)-3.\frac{4}{9}\)
=\(15.\frac{5}{9}-3.\frac{4}{9}\)
=\(\left[15-3.\left(\frac{5}{9}-\frac{4}{9}\right)\right]\)
=\(12.\frac{1}{9}\)
=\(\frac{4}{3}\)
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28 tháng 5 2022
a: \(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{13+\dfrac{13}{2}+\dfrac{13}{3}+\dfrac{13}{4}}{17-\dfrac{17}{2}+\dfrac{17}{3}-\dfrac{17}{4}}\)
\(=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}\cdot\dfrac{17\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{13\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)}=\dfrac{17}{13}\)
b: \(\dfrac{0.125-\dfrac{1}{5}+\dfrac{1}{7}}{0.375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0.2}{\dfrac{3}{4}+0.5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}=\dfrac{1}{3}+\dfrac{2}{3}=1\)