\(\frac{5}{6}+\frac{4}{5}+\frac{2}{5}\) ( tính giá trị biểu thức)
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a)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A = - 1\end{array}\)
b)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}\)
a) Với \(\frac{m}{n} = \frac{{ - 5}}{6}\), giá trị của biểu thức là:
\(\begin{array}{l}A = \frac{{ - 2}}{3} - \left( {\frac{{ - 5}}{6} + \frac{{ - 5}}{2}} \right).\frac{{ - 5}}{8}\\A = \frac{{ - 2}}{3} - \frac{{-20}}{6}.\frac{{ - 5}}{8}\\A = \frac{{ - 2}}{3} - \frac{{ 25}}{{12}}\\A = \frac{{ - 33}}{{12}}\end{array}\)
b) Với \(\frac{m}{n} = \frac{5}{2}\) , giá trị của biểu thức là:
\(\begin{array}{l}A = \frac{{ - 2}}{3} - \left( {\frac{5}{2} + \frac{{ - 5}}{2}} \right).\frac{{ - 5}}{8}\\A = \frac{{ - 2}}{3} - 0.\frac{{ - 5}}{8} = \frac{{ - 2}}{3}\end{array}\)
c) Với \(\frac{m}{n} = \frac{2}{{ - 5}}\) , giá trị của biểu thức là:
\(\begin{array}{l}A = \frac{-2}{3} - \left( {\frac{2}{{ - 5}} + \frac{{ - 5}}{2}} \right).\frac{{ - 5}}{8}\\A = \frac{-2}{3} - \left( {\frac{{ - 4}}{{10}} + \frac{{ - 25}}{{10}}} \right).\frac{{ - 5}}{8}\\A = \frac{-2}{3} - \frac{{ - 29}}{{10}}.\frac{{ - 5}}{8}\\A = \frac{-2}{3} - \frac{{29}}{{16}}\\A = \frac{{-32}}{{48}} - \frac{{87}}{{48}}\\A = \frac{{ - 119}}{{48}}\end{array}\).
a) Cách 1:
\(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}} + \frac{{ - 5}}{{ - 6}}} \right) + \frac{4}{5} = \frac{2}{5} + \frac{5}{6} + \frac{4}{5}\\ = \frac{{12}}{{30}} + \frac{{25}}{{30}} + \frac{{24}}{{30}} = \frac{{61}}{{30}}\end{array}\)
Cách 2:
\(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}} + \frac{{ - 5}}{{ - 6}}} \right) + \frac{4}{5} = \left( {\frac{2}{5} + \frac{4}{5}} \right) + \frac{5}{6}\\ = \frac{6}{5} + \frac{5}{6} = \frac{{36}}{{30}} + \frac{{25}}{{30}} = \frac{{61}}{{30}}\end{array}\)
b) Cách 1:
\(\begin{array}{l}\frac{{ - 3}}{{ - 4}} + \left( {\frac{{11}}{{ - 15}} + \frac{{ - 1}}{2}} \right) = \frac{3}{4} + \frac{{ - 11}}{{15}} + \frac{{ - 1}}{2}\\ = \frac{{45}}{{60}} + \frac{{ - 44}}{{60}} + \frac{{ - 30}}{{60}}\\ = \frac{{ - 29}}{{60}}\end{array}\).
Cách 2:
\(\begin{array}{l}\frac{{ - 3}}{{ - 4}} + \left( {\frac{{11}}{{ - 15}} + \frac{{ - 1}}{2}} \right) = \frac{3}{4} + \frac{{ - 11}}{{15}} + \frac{{ - 1}}{2}\\ = \left( {\frac{3}{4} + \frac{{ - 1}}{2}} \right) + \frac{{ - 11}}{{15}}\\ = \left( {\frac{3}{4} + \frac{{ - 2}}{4}} \right) + \frac{{ - 11}}{{15}}\\ = \frac{1}{4} + \frac{{ - 11}}{{15}}\\ = \frac{{15}}{{60}} + \frac{{ - 44}}{{60}}\\ = \frac{{ - 29}}{{60}}\end{array}\)
\(\frac{6:\frac{3}{5}-1\frac{1}{6}\cdot\frac{6}{7}}{4\frac{1}{5}\cdot\frac{10}{11}+5\frac{2}{11}}=1\)
\(\frac{6:\frac{3}{5}-1\frac{1}{6}\times\frac{6}{7}}{4\frac{1}{5}\times\frac{10}{11}+5\frac{2}{11}}\)
\(=\frac{\frac{6}{1}:\frac{3}{5}-\frac{7}{6}\times\frac{6}{7}}{\frac{21}{5}\times\frac{10}{11}\times\frac{57}{11}}\)
\(=\frac{\frac{6}{1}\times\frac{5}{3}-1}{\frac{210}{55}+\frac{57}{11}}\)
\(=\frac{\frac{30}{3}-1}{\frac{42}{11}+\frac{57}{11}}\)
\(=\frac{10-1}{\frac{99}{11}}\)
\(=\frac{9}{9}\)
\(=1\)
\(6:\frac{3}{5}-1\frac{1}{6}\)X \(\frac{6}{7}\) \(4\frac{1}{5}\)X \(\frac{10}{11}+5\frac{2}{11}\)
\(=\frac{33}{5}-\frac{7}{6}\)X \(\frac{6}{7}\) \(=\) \(\frac{21}{5}\)X \(\frac{10}{11}+\frac{57}{11}\)
\(=\frac{33}{5}-1\) \(=\frac{42}{11}+\frac{57}{11}\)
\(=\frac{28}{5}\) \(=\frac{99}{11}=9\)
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)
\(\left(6\frac{5}{5}-2\frac{4}{5}\right).\frac{25}{8}-\frac{8}{5}.\frac{1}{4}\)
\(=\frac{21}{5}.\frac{25}{8}-\frac{2}{5}\)
\(=\frac{105}{8}-\frac{2}{5}=\frac{525}{40}-\frac{16}{40}=\frac{509}{40}\)
\(\frac{5}{6}+\frac{4}{5}+\frac{2}{5}\)
\(=\frac{49}{30}+\frac{2}{5}\)
\(=\frac{61}{30}\)
\(\frac{5}{6}+\frac{4}{5}+\frac{2}{5}=\frac{5}{6}+\frac{6}{5}=\frac{61}{30}\)