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\(Q=\left(\frac{-9}{25}\right).\frac{53}{3}-\left(\frac{-3}{5}\right)^2.\frac{22}{3}\)
\(Q=\left(\frac{-9}{25}\right).\frac{53}{3}-\left(\frac{9}{25}\right).\frac{22}{3}\)
\(Q=\left(\frac{9}{25}\right).\frac{-53}{3}-\left(\frac{9}{25}\right).\frac{22}{3}\)
\(Q=\frac{9}{25}.\left(\frac{-53}{3}-\frac{22}{3}\right)\)
\(Q=\frac{9}{25}.\frac{-75}{3}\)
\(Q=\frac{9}{25}.\left(-25\right)\)
\(Q=\frac{-225}{25}=-9\)
Q= \(\left(\frac{-9}{25}\right).\frac{53}{3}-\left(\frac{-3}{5}\right)^2.\frac{22}{3}\)
Q=\(\left(\frac{-9}{25}\right).\frac{53}{3}-\frac{9}{25}.\frac{22}{3}\)
Q=\(\frac{9}{25}.\left(\frac{-53}{3}\right)-\frac{9}{25}.\frac{22}{3}\)
Q=\(\frac{9}{25}.\left(\frac{-53}{3}-\frac{22}{3}\right)\)
Q=\(\frac{9}{25}.\left(-25\right)\)=\(-9\)
a) \(\left( {\frac{7}{3} + 3,5} \right):\left( { - \frac{{25}}{6} + \frac{{22}}{7}} \right) + 0,5\)
\(\begin{array}{l} = \left( {\frac{7}{3} + \frac{7}{2}} \right):\left( { - \frac{{25}}{6} + \frac{{22}}{7}} \right) + \frac{1}{2}\\ = \frac{{35}}{6}:\frac{{ - 25.7 + 22.6}}{{6.7}} + \frac{1}{2}\\ = \frac{{35}}{6}:\frac{{ - 43}}{{7.6}} + \frac{1}{2} = \frac{{35}}{6}.\frac{{7.6}}{{ - 43}} + \frac{1}{2}\\ = \frac{{ - 245}}{{43}} + \frac{1}{2} = \frac{{ - 245.2 + 43}}{{43.2}} = \frac{{ - 447}}{{86}}\end{array}\)
b) \(\frac{{38}}{7} + \left( { - 3,25} \right) - \frac{{17}}{7} + 4,55\)
\(\begin{array}{l} = \left( {\frac{{38}}{7} - \frac{{17}}{7}} \right) + \left( {4,55 - 3,25} \right)\\ = \frac{{38 - 17}}{7} + 1,3 = \frac{{21}}{7} +1,3\\ = 3 + 1,3 = 4,3\end{array}\)
\(-\frac{9}{11}\cdot\frac{3}{8}-\frac{9}{11}\cdot\frac{5}{8}+\frac{17}{11}=-\frac{9}{11}\left(\frac{3}{8}+\frac{5}{8}\right)+\frac{17}{11}=-\frac{9}{11}\cdot1+\frac{17}{11}=1\)
\(\frac{2}{1.3}+....+\frac{2}{53.55}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{53}-\frac{1}{55}=1-\frac{1}{55}=\frac{54}{55}\)
\(x+5-\frac{1}{2}=3\frac{1}{2}\)
\(x+5=3.5+0.5=4\)
\(x=4-5=-1\)
\(3^{x+1}=27=3^3\)
\(x+1=3\)
vậy x=2
a,\(\frac{5}{3}.\frac{3}{7}+\frac{5}{3}.\frac{5}{7}-\frac{5}{3}\)
=\(\frac{5}{3}.\left(\frac{3}{7}+\frac{5}{7}\right)-\frac{5}{3}\)
= \(\frac{5}{21}\)
\(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}\) _ \(\frac{3-\frac{3}{11}-\frac{3}{17}}{5-\frac{5}{11}-\frac{5}{17}}\)
=\(\frac{2\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}{4\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}\)_ \(\frac{3\left(1-\frac{1}{11}-\frac{1}{17}\right)}{5\left(1-\frac{1}{11}-\frac{1}{17}\right)}\)= \(\frac{2}{4}-\frac{3}{5}\)= \(\frac{-1}{10}\)
`Answer:`
\(-\frac{9}{25}.17\frac{2}{3}-\left(-\frac{3}{5}\right)^2.\frac{22}{3}\)
\(=-\frac{9}{25}.\frac{53}{3}-\frac{9}{25}.\frac{22}{3}\)
\(=\frac{9}{25}.-\frac{53}{3}-\frac{9}{25}.\frac{22}{3}\)
\(=\frac{9}{25}.\left(-\frac{53}{3}-\frac{22}{3}\right)\)
\(=\frac{9}{25}.-\frac{75}{3}\)
\(=-9\)
\(=-\left(\dfrac{3}{5}\right)^2\times\dfrac{53}{3}-\left(\dfrac{3}{5}\right)^2\times\dfrac{22}{3}\)
\(=\left(\dfrac{3}{5}\right)^2\left(-\dfrac{53}{3}-\dfrac{23}{3}\right)\)
\(=\dfrac{9}{25}\times\left(-\dfrac{76}{3}\right)=-\dfrac{228}{25}\)