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a. \(\frac{-3}{2}-2x+\frac{3}{4}=-22\)2
=> \(-2x=-22+\frac{3}{2}-\frac{3}{4}\)
=> \(-2x=\frac{-85}{4}\)
=> \(x=\frac{-85}{4}:\left(-2\right)\)
=> \(x=\frac{85}{8}\)
b. \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\left(\frac{3}{-2}-\frac{10}{3}\right)=\frac{2}{5}\)
=> \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\frac{-29}{6}=\frac{2}{5}\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{2}{5}:\left(\frac{-29}{6}\right)\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{-12}{145}\)
=> \(\frac{-2}{3}x=\frac{-12}{145}+\frac{3}{5}\)
=> \(\frac{-2}{3}x=\frac{15}{29}\)
=> x = \(\frac{15}{29}:\frac{-2}{3}\)
=> x = \(\frac{-45}{58}\)
\(\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}=-\frac{19}{20}.\frac{5}{19}=-\frac{1}{4}\)
\(\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}=\frac{3}{10}.-\frac{4}{3}=-\frac{4}{10}=-\frac{2}{5}\)
\(\dfrac{3}{4}\times\dfrac{8}{5}:1\dfrac{1}{6}\)
=\(\dfrac{6}{5}:\) \(\dfrac{7}{6}\)
=\(\dfrac{6}{5}\times\dfrac{6}{7}=\dfrac{36}{35}\)
2\(\dfrac{1}{3}\) x 1\(\dfrac{1}{4}\) -\(\dfrac{7}{5}\)
\(\dfrac{7}{3}\times\dfrac{5}{4}-\) \(\dfrac{7}{5}\)
\(\dfrac{35}{12}-\dfrac{7}{5}\)
\(\dfrac{175}{60}-\dfrac{84}{60}=\dfrac{91}{60}\)
4\(\dfrac{2}{3}+1\dfrac{1}{4} +2\dfrac{1}{3}+2\dfrac{3}{7}\)
(4 +2) + \(\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\) +1\(\dfrac{1}{4}\) + \(2\dfrac{3}{7}\)
6 + 1 + \(\dfrac{5}{4}\) + \(\dfrac{17}{7}\)
7 + \(\dfrac{103}{28}\)
\(\dfrac{299}{28}\)
\(\frac{\left(\frac{3}{15}+\frac{1}{4}+\frac{7}{20}\right)\times\frac{17}{49}}{5\frac{1}{3}+\frac{2}{5}}=\frac{\left(\frac{1}{5}+\frac{1}{4}+\frac{7}{20}\right)\times\frac{17}{49}}{\frac{16}{3}+\frac{2}{5}}\)
=\(\frac{\left(\frac{4}{20}+\frac{5}{20}+\frac{7}{20}\right)\times\frac{17}{49}}{\frac{80}{15}+\frac{6}{15}}=\frac{\frac{16}{20}\times\frac{17}{49}}{\frac{86}{15}}=\frac{\frac{4}{5}\times\frac{17}{49}}{\frac{86}{15}}\)
=\(\frac{68}{245}\times\frac{15}{86}=\frac{102}{2107}\)
\(\frac{3}{5}-\frac{1}{15}.\frac{10}{13}=\frac{3}{5}-\frac{2}{39}=\frac{107}{195}\)
\(\frac{1}{2}+\frac{3}{8}:\frac{3}{4}=\frac{1}{2}+\frac{3}{8}.\frac{4}{3}=\frac{1}{2}+\frac{1}{2}=1\)
\(\left(\frac{3}{5}-\frac{3}{20}\right):\frac{4}{5}=\frac{9}{20}.\frac{5}{4}=\frac{9}{16}\)
\(\frac{107}{195}\) 1 \(\frac{9}{16}\)