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\(a)\frac{2}{{15}} + \left( {\frac{{ - 5}}{{24}}} \right) = \frac{{16}}{{120}} + \left( {\frac{{ - 25}}{{120}}} \right) = \frac{{ - 9}}{{120}} = \frac{{ - 3}}{{40}}\)
b) \(\left( {\frac{{ - 5}}{9}} \right) - \left( { - \frac{7}{{27}}} \right) = \left( {\frac{{ - 15}}{{27}}} \right) + \frac{7}{{27}} = \frac{{ - 8}}{{27}}\)
c)\(\left( { - \frac{7}{{12}}} \right) + 0,75 = \left( { - \frac{7}{{12}}} \right) + \frac{75}{100} \\= \left( { - \frac{7}{{12}}} \right) + \frac{3}{4} \\= \left( { - \frac{7}{{12}}} \right) + \frac{9}{{12}} = \frac{2}{{12}} = \frac{1}{6}\)
d)\(\left( {\frac{{ - 5}}{9}} \right) - 1,25 =\left( {\frac{{ - 5}}{9}} \right) - \frac{125}{100} = \left( {\frac{{ - 5}}{9}} \right) - \frac{5}{4}\\ = \left( {\frac{{ - 20}}{{36}}} \right) - \frac{{45}}{{36}} = \frac{{ - 65}}{{36}}\)
e)\(0,34.\frac{{ - 5}}{{17}} =\frac{{34}}{{100}}.\frac{{ - 5}}{{17}} = \frac{{17}}{{50}}.\frac{{ - 5}}{{17}} = \frac{{ - 1}}{{10}}\)
g) \(\frac{4}{9}:\left( { - \frac{8}{{15}}} \right) = \frac{4}{9}.\left( { - \frac{{15}}{8}} \right) = \frac{{ - 5}}{6}\)
h)\(\left( {1\frac{2}{3}} \right):\left( {2\frac{1}{2}} \right) = \frac{5}{3}:\frac{5}{2} = \frac{5}{3}.\frac{2}{5} = \frac{2}{3}\)
i) \(\frac{2}{5}.\left( { - 1,25} \right) = \frac{2}{5}.\frac{{ - 125}}{100} = \frac{2}{5}.\frac{{ - 5}}{4} = \frac{{ - 1}}{2}\)
k) \(\left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).3\frac{1}{9} = \left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).\frac{{28}}{9}\\ = \frac{{ - 3.3.5.7.4}}{{5.\left( { - 7} \right).3.3}} = 4\)
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
Bài 1:
a) Ta có: \(25\cdot\left(\frac{-1}{5}\right)^3+\frac{1}{5}-2\cdot\left(\frac{-1}{2}\right)^2-\frac{1}{2}\)
\(=25\cdot\frac{-1}{125}+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\)
\(=-\frac{1}{5}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}\)
\(=\frac{-2}{2}=-1\)
b) Ta có: \(35\frac{1}{6}:\left(\frac{-4}{5}\right)-46\frac{1}{6}:\left(\frac{-4}{5}\right)\)
\(=\frac{211}{6}\cdot\frac{-5}{4}-\frac{277}{6}\cdot\frac{-5}{4}\)
\(=\frac{-5}{4}\cdot\left(\frac{211}{6}-\frac{277}{6}\right)\)
\(=\frac{-5}{4}\cdot\left(-11\right)=\frac{55}{4}\)
c) Ta có: \(\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=\frac{-7}{20}\cdot\frac{7}{3}+\frac{7}{20}\cdot\frac{7}{3}\)
\(=\frac{7}{3}\cdot\left(-\frac{7}{20}+\frac{7}{20}\right)=\frac{7}{3}\cdot0=0\)
d) Ta có: \(\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}\cdot\left(\frac{1}{36}-\frac{5}{12}\right)\)
\(=\frac{7}{8}\cdot6+\frac{7}{8}\cdot\frac{-7}{18}\)
\(=\frac{7}{8}\cdot\left(6+\frac{-7}{18}\right)\)
\(=\frac{7}{8}\cdot\frac{101}{18}=\frac{707}{144}\)
e) Ta có: \(\frac{1}{6}+\frac{5}{6}\cdot\frac{3}{2}-\frac{3}{2}+1\)
\(=\frac{1}{6}+\frac{15}{12}-\frac{3}{2}+1\)
\(=\frac{2}{12}+\frac{15}{12}-\frac{18}{12}+\frac{12}{12}\)
\(=\frac{11}{12}\)
f) Ta có: \(\left(-0,75-\frac{1}{4}\right):\left(-5\right)+\frac{1}{15}-\left(-\frac{1}{5}\right):\left(-3\right)\)
\(=\left(-1\right):\left(-5\right)+\frac{1}{15}-\frac{1}{15}\)
\(=\frac{1}{5}\)
c) C = ( 1 - 2 ) + ( 3 - 4 ) + ... + ( 79 - 80 )
C = ( -1 ) + ( -1 ) + ... + ( -1 )
C = ( -1 ) x ( 80 - 1 + 1 ) : 2
C = ( -1 ) x 80 : 2
C = ( -40 )
a)
\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)
Vậy \(x = \frac{{ - 2}}{3}\).
b)
\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)
Vậy\(x = \frac{1}{12}\).
c)
\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)
Vậy \(x = \frac{7}{3}\).
d)
\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)
Vậy \(x = \frac{{ - 9}}{{10}}\).
a)\({\left( {\frac{2}{5} + \frac{1}{2}} \right)^2} = {\left( {\frac{4}{{10}} + \frac{5}{{10}}} \right)^2} = {\left( {\frac{9}{{10}}} \right)^2} = \frac{{81}}{{100}}\);
b)\({\left( {0,75 - 1\frac{1}{2}} \right)^3} = {\left( {\frac{3}{4} - \frac{3}{2}} \right)^3} = {\left( {\frac{3}{4} - \frac{6}{4}} \right)^3} = {\left( { - \frac{3}{4}} \right)^3} = \frac{{ - 27}}{{64}};\)
c)
\(\begin{array}{l}{\left( {\frac{3}{5}} \right)^{15}}:{\left( {0,36} \right)^5} = {\left( {\frac{3}{5}} \right)^{15}}:{\left( {\frac{9}{{25}}} \right)^5}\\ = {\left( {\frac{3}{5}} \right)^{15}}:{\left[ {{{\left( {\frac{3}{5}} \right)}^2}} \right]^5} = {\left( {\frac{3}{5}} \right)^{15}}:{\left( {\frac{3}{5}} \right)^{10}} = {\left( {\frac{3}{5}} \right)^5}\end{array}\)
d) \({\left( {1 - \frac{1}{3}} \right)^8}:{\left( {\frac{4}{9}} \right)^3} = {\left( {\frac{3}{3} - \frac{1}{3}} \right)^8}:{\left( ({\frac{2}{3}})^2 \right)^3}\\= {\left( {\frac{2}{3}} \right)^8}:{\left( {\frac{2}{3}} \right)^6} = {\left( {\frac{2}{3}} \right)^{8-6}}\\= {\left( {\frac{2}{3}} \right)^2} = \frac{4}{9}\)
\(\frac{\left(\frac{1}{2}-\frac{3}{4}\right).\left(\frac{1}{5}-\frac{2}{5}\right)}{\frac{5}{9}-\frac{13}{12}}=-\frac{\frac{1}{4}.\left(-\frac{1}{5}\right)}{-\frac{19}{36}}=-\frac{9}{95}\)