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\(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)\)
\(=\frac{5}{9}:\frac{-3}{22}+\frac{5}{9}:\frac{-3}{5}\)
\(=\frac{5}{9}\cdot\frac{-22}{3}+\frac{5}{9}\cdot\frac{-5}{3}\)
\(=\frac{5}{9}\cdot\left(\frac{-22}{3}+\frac{-5}{3}\right)\)
\(=\frac{5}{9}\cdot\left(-9\right)\)
\(=-5\)
\(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)\)
\(=\frac{5}{9}:\left(\frac{2}{22}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{10}{15}\right)\)
\(=\frac{5}{9}:\frac{\left(-3\right)}{22}+\frac{5}{9}:\frac{\left(-9\right)}{15}\)
\(=\frac{5}{9}.\frac{22}{\left(-3\right)}+\frac{5}{9}.\frac{15}{\left(-9\right)}\)
\(=\frac{5}{9}.\left(\frac{-22}{3}+\frac{-15}{9}\right)\)
\(=\frac{5}{9}.\left(\frac{-66}{9}+\frac{-15}{9}\right)\)
\(=\frac{5}{9}.\frac{\left(-81\right)}{9}=-5\)
\(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)\)
\(=\frac{5}{9}:\left(\frac{-3}{22}\right)+\frac{5}{9}:\left(\frac{-3}{5}\right)\)
\(=\frac{5}{9}.\left(-\frac{22}{3}\right)+\frac{5}{9}.\left(-\frac{5}{3}\right)\)
\(=\frac{5}{9}.\left[-\frac{22}{3}+\left(-\frac{5}{3}\right)\right]\)
\(=\frac{5}{9}.\left(-9\right)\)
\(=-5\)
\(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)\)
\(=\frac{5}{9}:\left[\left(\frac{1}{11}-\frac{5}{22}\right)+\left(\frac{1}{15}-\frac{2}{3}\right)\right]\)
\(=\frac{5}{9}:\left[-\frac{3}{22}-\frac{3}{5}\right]\)
\(=\frac{5}{9}:\frac{-81}{110}=\frac{5}{9}.\frac{-110}{81}\)
\(=-\frac{550}{729}\)
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
Dùng tích chất kết hợp cho nó lẹ
a/\(\left(\frac{-2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(\frac{-1}{3}+\frac{4}{7}\right):\frac{4}{5}=\left(\frac{-2}{3}+\frac{3}{7}+\frac{-1}{3}+\frac{4}{7}\right):\frac{4}{5}=\left(-1+1\right):\frac{4}{5}=0\)
b/\(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)=\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}+\frac{1}{15}-\frac{2}{3}\right)=\frac{5}{9}:\left(\frac{-3}{22}+\frac{-3}{5}\right)=\frac{-5}{3\left(\frac{1}{22}+\frac{1}{5}\right)}=\frac{-550}{81}\)
Mà hình như câu b mình làm sai
b/\(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)=\frac{5}{9}:\frac{-3}{22}+\frac{5}{9}:\frac{-3}{5}=\frac{5.22}{9.-3}+\frac{5.5}{9.-3}=\frac{-\left(5.22+5.5\right)}{27}=-5\)
Đúng rồi em ạ :)
đúng đó nhưng bạn nên nhóm \(\frac{-22}{3}\)và \(\frac{-5}{3}\)vào với nhau và đặt \(\frac{5}{9}\)ra ngoài sẽ tính nhanh hơn