Nghịch đảo của kết quả phép tính :\(5-\frac{5}{7-\frac{5}{7-\frac{5}{7-\frac{5}{7}}}}\)
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Mình làm như thế này nek
\(\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{0,75+\frac{9}{7}-2\frac{2}{5}}+\frac{\frac{3}{14}-\frac{2}{10}+\frac{5}{18}+\frac{7}{66}}{\frac{6}{7}-\frac{4}{5}+\frac{10}{9}+\frac{14}{33}}\)
\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{\frac{2}{4}+\frac{9}{7}-\frac{12}{5}}+\frac{\frac{1}{2}\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}{2\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}\)
\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{3\cdot\left(\frac{1}{4}+\frac{3}{7}-\frac{4}{5}\right)}+\frac{\frac{1}{2}}{2}\)
\(=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
a) \(\frac{6}{5}.{\left( {1,2} \right)^8} = 1,2.{(1,2)^8} = {(1,2)^{1 + 8}} = {(1,2)^9}\)
b) \({\left( {\frac{{ - 4}}{9}} \right)^7}:\frac{{16}}{{81}} = {\left( {\frac{{ - 4}}{9}} \right)^7}:{\left( {\frac{{ - 4}}{9}} \right)^2} = {\left( {\frac{{ - 4}}{9}} \right)^{7 - 2}} = {\left( {\frac{{ - 4}}{9}} \right)^5}\)
a)
\(\begin{array}{l}\frac{2}{3} + \frac{{ - 2}}{5} + \frac{{ - 5}}{6} - \frac{{13}}{{10}}\\ = \frac{2}{3} + \frac{{ - 5}}{6} + \frac{{ - 2}}{5} - \frac{{13}}{{10}}\\ = \left( {\frac{2}{3} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 2}}{5} - \frac{{13}}{{10}}} \right)\\ = \left( {\frac{4}{6} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 4}}{{10}} - \frac{{13}}{{10}}} \right)\\ = \frac{{ - 1}}{6} + \frac{{ - 17}}{{10}}\\ = \frac{{ - 5}}{{30}} + \frac{{ - 51}}{{30}}\\ = \frac{{ - 56}}{{30}}\\ = \frac{{ - 28}}{{15}}\end{array}\)
b)
\(\begin{array}{l}\frac{{ - 3}}{7}.\frac{{ - 1}}{9} + \frac{7}{{ - 18}}.\frac{{ - 3}}{7} + \frac{5}{6}.\frac{{ - 3}}{7}\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 1}}{9} + \frac{7}{{ - 18}} + \frac{5}{6}} \right)\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 7}}{{18}} + \frac{{15}}{{18}}} \right)\\ = \frac{{ - 3}}{7}.\frac{{ 6}}{{18}}\\ = \frac{-1}{7}\end{array}\).
\(\dfrac{3}{5}:\left(\dfrac{1}{4}.\dfrac{7}{5}\right)=\dfrac{3}{5}:\dfrac{7}{20}\)
\(=\dfrac{60}{35}=\dfrac{12}{7}\)
=> D
Câu 3 : \(\dfrac{-6}{11}=>\dfrac{-11}{6}\)
a)
\(\begin{array}{l}M = \frac{1}{7}.(\frac{{ - 5}}{8}) + \frac{1}{7}.(\frac{{ - 11}}{8})\\ = \frac{{ - 5}}{{56}} + \frac{{ - 11}}{{56}} = \frac{{ - 16}}{{56}} = \frac{{ - 2}}{7}\end{array}\)
b)
\(\begin{array}{l}M = \frac{1}{7}.(\frac{{ - 5}}{8}) + \frac{1}{7}.(\frac{{ - 11}}{8})\\ = \frac{1}{7}.[(\frac{{ - 5}}{8}) + (\frac{{ - 11}}{8})]\\ = \frac{1}{7}.\frac{{ - 16}}{8}\\ = \frac{1}{7}.( - 2)\\ = \frac{{ - 2}}{7}\end{array}\)
\(\frac{3}{5}:\frac{7}{3}+\frac{3}{5}:\frac{7}{4}-1\frac{3}{5}\)
\(=\frac{3}{5}.\frac{3}{7}+\frac{3}{5}.\frac{4}{7}-1\frac{3}{5}\)
\(=\frac{3}{5}.\left(\frac{3}{7}+\frac{4}{7}\right)-1\frac{3}{5}\)
\(=\frac{3}{5}.1-1+\frac{3}{5}\)
\(=\frac{3}{5}-\frac{3}{5}-1\)
\(=-1\)
a) 35 :73 +35 :74 −135
\(=\frac{3}{5}:\left(\frac{7}{3}+\frac{7}{4}\right)-\frac{8}{5}\) \(=\frac{3}{5}.\frac{49}{12}-\frac{8}{5}\)\(=\frac{49}{20}-\frac{8}{5}\)
= \(\frac{17}{20}\)
Có : \(5-\frac{5}{7-\frac{5}{7-\frac{5}{\frac{44}{7}}}}=5-\frac{5}{7-\frac{5}{7-5:\frac{44}{7}}}=5-\frac{5}{7-\frac{5}{7-\frac{35}{44}}}\)
\(=5-\frac{5}{7-\frac{5}{\frac{273}{44}}}=5-\frac{5}{7-5:\frac{273}{44}}=5-\frac{5}{7-\frac{220}{273}}\)
\(=5-\frac{5}{\frac{1691}{273}}=5-5:\frac{1691}{273}=5-\frac{1365}{1691}=\frac{7090}{1691}\)
\(\Rightarrow\)Số nghịch đảo của biểu thức trên là \(\frac{1691}{7090}\)
\(5-\frac{5}{7-\frac{5}{7-\frac{5}{7-\frac{5}{7}}}}\) \(\Rightarrow5-\frac{5}{7-\frac{5}{7-\frac{5}{\frac{44}{7}}}}=5-\frac{5}{7-\frac{5}{7-\frac{35}{44}}}=5-\frac{5}{7-\frac{5}{\frac{273}{44}}}=5-\frac{5}{7-\frac{220}{273}}\)
\(5-\frac{5}{\frac{1691}{273}}=5-\frac{1365}{1691}=\frac{7090}{1691}\)