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3) C thiếu đề
4) \(D=\frac{1}{9}-\left|\frac{-5}{23}\right|-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{5}{23}+\frac{5}{23}-\frac{1}{9}-\frac{25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{1}{9}-\frac{5}{23}+\frac{5}{23}+\frac{-25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=0+0+\frac{125}{14}-\frac{7}{30}\)
\(D=\frac{913}{105}\)
\(\frac{\frac{-4}{5}+\frac{4}{19}-\frac{4}{23}}{\frac{8}{5}-\frac{8}{19}+\frac{8}{23}}=\left(-\frac{4}{5}+\frac{4}{19}-\frac{4}{23}\right)\cdot\frac{1}{\frac{8}{5}-\frac{8}{19}+\frac{8}{23}}=\left(-\frac{4}{5}+\frac{4}{19}-\frac{4}{23}\right)\cdot\frac{1}{-2\cdot\left(-\frac{4}{5}+\frac{4}{19}-\frac{4}{23}\right)}=-\frac{1}{2}\)
Ta có : \(A=\frac{-\frac{4}{9}+\frac{4}{19}-\frac{4}{23}}{\frac{8}{5}-\frac{8}{19}+\frac{8}{23}}\)
\(\Leftrightarrow A=\frac{-\left(\frac{4}{9}-\frac{4}{19}+\frac{4}{23}\right)}{\frac{8}{5}-\frac{8}{19}+\frac{8}{23}}\)
\(\Leftrightarrow A=\frac{-4\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{23}\right)}{8\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}\)
\(\Leftrightarrow A=-\frac{4}{8}=-\frac{1}{2}\)
\(A=-\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
\(=\frac{-6.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}\)
\(=-\frac{6}{9}=-\frac{2}{3}\)
a) Cách 1:
\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = 8 + \frac{7}{3} - \frac{3}{5} - 5 - \frac{2}{5} - \frac{{10}}{3} + 2\\ = (8 - 5 + 2) + (\frac{7}{3} - \frac{{10}}{3}) - (\frac{3}{5} + \frac{2}{5})\\ = 5 + \frac{{ - 3}}{3} - \frac{5}{5}\\ = 5 + ( - 1) - 1\\ = 3\end{array}\)
Cách 2:
\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = (\frac{{120}}{{15}} + \frac{{35}}{{15}} - \frac{9}{{15}}) - (\frac{{25}}{5} + \frac{2}{5}) - (\frac{{10}}{3} - \frac{6}{3})\\ = \frac{{146}}{{15}} - \frac{{27}}{5} - \frac{4}{3}\\ = \frac{{146}}{{15}} - \frac{{81}}{{15}} - \frac{{20}}{{15}}\\ = \frac{{45}}{{15}}\\ = 3\end{array}\)
b)
\(\begin{array}{l}(7 - \frac{1}{2} - \frac{3}{4}):(5 - \frac{1}{4} - \frac{5}{8})\\ = (\frac{{28}}{4} - \frac{2}{4} - \frac{3}{4}):(\frac{{40}}{8} - \frac{2}{8} - \frac{5}{8})\\ = \frac{{23}}{4}:\frac{{33}}{8}\\ = \frac{{23}}{4}.\frac{8}{{33}}\\ = \frac{{46}}{{33}}\end{array}\)
\(H=\frac{4}{15}-\frac{23}{28}-\left(-\frac{23}{28}+\frac{-11}{15}-\frac{24}{27}\right)-\frac{2}{27}\)
\(H=\frac{4}{15}+\frac{-23}{28}+\frac{23}{28}+\frac{11}{15}+\frac{24}{27}+\frac{-2}{27}\)
\(H=\left(\frac{4}{15}+\frac{11}{15}\right)+\left(\frac{-23}{28}+\frac{23}{28}\right)+\left(\frac{24}{27}+\frac{-2}{27}\right)\)
\(H=1+0+\frac{22}{27}\)
\(H=1+\frac{22}{27}\)
\(H=\frac{27}{27}+\frac{22}{27}=\frac{49}{27}\)
\(H=\frac{4}{15}-\frac{23}{28}-\left(\frac{-23}{28}+\frac{-11}{15}-\frac{24}{27}\right)-\frac{2}{27}\)
\(H=\frac{4}{15}-\frac{23}{28}+\frac{23}{28}+\frac{11}{15}+\frac{24}{27}-\frac{2}{27}\)
\(H=\left(\frac{4}{15}+\frac{11}{15}\right)+\left(\frac{-23}{28}+\frac{23}{28}\right)+\left(\frac{24}{27}-\frac{2}{27}\right)\)
\(H=1+0+\frac{22}{27}\)
\(H=\frac{49}{27}\)
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)
\(A=\frac{\frac{-4}{5}+\frac{4}{19}-\frac{4}{23}}{\frac{8}{5}-\frac{8}{19}+\frac{8}{23}}=\frac{-\left(\frac{4}{5}-\frac{4}{19}+\frac{4}{23}\right)}{2\left(\frac{4}{5}-\frac{4}{19}+\frac{4}{23}\right)}=-\frac{1}{2}\)