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\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)
= \(\frac{\left(\frac{1}{2}\right)^2\cdot2018-\left(\left(\frac{1}{2}\right)^2\right)^6\cdot2017}{\left(\frac{1}{2}\right)^2\cdot\frac{1}{3}\cdot\left(\frac{1}{2}\right)^{13}}\)
= \(\frac{\left(\frac{1}{2}\right)^2\cdot2018-\left(\frac{1}{2}\right)^{12}\cdot2017}{\left(\frac{1}{2}\right)^{15}\cdot\frac{1}{3}}\)
=\(\frac{\left(\frac{1}{2}\right)^2\cdot\left(2018-2017\right)\cdot\left(\frac{1}{2}\right)^{10}}{\left(\frac{1}{2}\right)^{15}.\frac{1}{3}}\)
= \(\frac{\left(\frac{1}{2}\right)^2\cdot1\cdot\left(\frac{1}{2}\right)^{10}}{\left(\frac{1}{2}\right)^{15}\cdot\frac{1}{3}}\)
= \(\frac{\left(\frac{1}{2}\right)^{12}}{\left(\frac{1}{2}\right)^{15}\cdot\frac{1}{3}}\)
= \(\frac{1}{\left(\frac{1}{2}\right)^3\cdot\frac{1}{3}}\)
= \(\frac{1}{\frac{1}{24}}\)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)