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\(B=\left(8-2,25+\dfrac{2}{7}\right)-\left(-6-\dfrac{3}{7}+1\dfrac{1}{4}\right)-\left(3+0,5-1\dfrac{2}{7}\right)\\ B=8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{1}{2}+\dfrac{9}{7}\\ B=\left(8+6-3\right)+\left(-\dfrac{9}{4}-\dfrac{5}{4}\right)+\left(\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}\right)-\dfrac{1}{2}\)
\(B=11-\dfrac{7}{2}+2-\dfrac{1}{2}\\ B=\left(11+2\right)+\left(-\dfrac{7}{2}-\dfrac{1}{2}\right)\\ B=13-4\\ B=9.\)
\(\left[-\sqrt{2,25}+4\sqrt{\left(-2,15\right)^2}-\left(3\sqrt{\dfrac{7}{6}}\right)^2\right]\sqrt{1\dfrac{9}{16}}\)
\(=\left[-1,5+4\sqrt{2,15^2}-9\cdot\dfrac{7}{6}\right]\sqrt{\dfrac{25}{16}}\)
\(=\left[4\cdot\dfrac{43}{20}-10,5-1,5\right]\cdot\dfrac{5}{4}\)
\(=\left[\dfrac{43}{5}-12\right]\cdot\dfrac{5}{4}\)
\(=\dfrac{43}{5}\cdot\dfrac{5}{4}-12\cdot\dfrac{5}{4}\)
\(=\dfrac{43}{4}-15=\dfrac{-17}{4}\)
a) \(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{7}{4}:\frac{2}{5}\)
\(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{35}{8}\)
\(\Rightarrow\frac{1}{3}.x=\frac{35}{8}.\frac{2}{3}\)
\(\Rightarrow\frac{1}{3}.x=\frac{35}{12}\)
\(\Rightarrow x=\frac{35}{12}:\frac{1}{3}\)
\(\Rightarrow x=\frac{35}{4}\)
Vậy \(x=\frac{35}{4}\)
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
5: \(=3-\dfrac{1}{4}+\dfrac{2}{3}-5+\dfrac{1}{3}+\dfrac{6}{5}-6+\dfrac{7}{4}-\dfrac{3}{2}\)
\(=3-5-6+\dfrac{-1}{4}+\dfrac{7}{4}+\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{6}{5}-\dfrac{3}{2}\)
\(=-8+\dfrac{3}{2}+1+\dfrac{-3}{10}\)
\(=-7+\dfrac{15-3}{10}=-7+\dfrac{6}{5}=-\dfrac{29}{5}\)
6: \(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=6-5-3-\dfrac{2}{3}-\dfrac{5}{3}+\dfrac{7}{3}+\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\)
\(=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)
7: \(=\dfrac{5}{3}-\dfrac{3}{7}+9-2-\dfrac{5}{7}+\dfrac{2}{3}+\dfrac{8}{7}-\dfrac{4}{3}-10\)
\(=9-2-10+\dfrac{5}{3}+\dfrac{2}{3}-\dfrac{4}{3}+\dfrac{-3}{7}-\dfrac{5}{7}+\dfrac{8}{7}\)
=-3+1
=-2
8: \(=8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{2}{4}+\dfrac{9}{7}\)
\(=8+6-3+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}-1-\dfrac{2}{4}\)
\(=11+2-1-\dfrac{1}{2}\)
=11+1/2
=11,5
- \(\dfrac{3}{7}\) + (3 - \(\dfrac{3}{4}\)) - (2,25 - \(\dfrac{10}{7}\))
= - \(\dfrac{3}{7}\) + 3 - 0,75 - 2,25 + \(\dfrac{10}{7}\)
= ( - \(\dfrac{3}{7}\) + \(\dfrac{10}{7}\)) + 3 - (0,75 + 2,25)
= 1 + 3 - 3
= 1