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\(P=\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):\left(-2\right)\)
\(P=\left(-1,1\right):\left(-3\right)+\frac{1}{3}+\frac{1}{6}:\left(-2\right)\)
\(P=\frac{11}{30}+\frac{1}{3}+\left(-\frac{1}{12}\right)\)
\(P=\frac{37}{60}\)
\(Q=\left(\frac{2}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right).2\frac{2}{17}\right]\)
\(Q=\left(-0,928\right):\frac{4}{7}:\left[\left(-\frac{119}{36}\right).2\frac{2}{17}\right]\)
\(Q=\left(-1,624\right):\left(-\frac{245}{36}\right)\)
\(Q=\frac{1044}{4375}\)
\(=\frac{-\frac{1}{8}-\frac{27}{64}.4}{-2+\frac{9}{16}-\frac{3}{8}}\)
\(=\frac{-\frac{1}{8}-\frac{27}{16.4}.4}{-2+\frac{9-6}{16}}\)
\(=\frac{-\frac{1}{8}-\frac{27}{16}}{-2+\frac{3}{16}}\)
\(=\frac{-\left(\frac{2+27}{16}\right)}{\frac{-32+3}{16}}\)
\(=\frac{-\frac{29}{16}}{\frac{-29}{16}}\)
\(=1\)
\(P=\left(\dfrac{-1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{6}:2\)
\(=\left(\dfrac{1}{2}+\dfrac{3}{5}\right):3+\dfrac{1}{3}-\dfrac{1}{12}\)
\(=\dfrac{11}{10}\cdot\dfrac{1}{3}+\dfrac{1}{4}\)
\(=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{22}{60}+\dfrac{15}{60}=\dfrac{37}{60}\)
\(Q=\left(\dfrac{2}{25}-\dfrac{126}{125}\right)\cdot\dfrac{7}{4}:\left[\dfrac{-119}{36}\cdot\dfrac{36}{17}\right]\)
\(=\dfrac{-116}{125}\cdot\dfrac{7}{4}:\left(-7\right)\)
\(=\dfrac{116}{125}\cdot\dfrac{7}{4}\cdot\dfrac{1}{7}=\dfrac{29}{125}\)
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)
a)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A = - 1\end{array}\)
b)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}\)
Ta có:
\(S=\left(\frac{3}{2}-\frac{2}{2^2}\right)\left(\frac{4}{3}-\frac{2}{3^2}\right)\left(\frac{5}{4}-\frac{2}{4^2}\right)...\left(\frac{101}{100}-\frac{2}{100^2}\right)\)
\(=\frac{4}{2^2}.\frac{10}{3^2}.\frac{18}{4^2}....\frac{100.101-2}{101^2}\)
\(=\frac{1.4}{2^2}.\frac{2.5}{3^2}.\frac{3.6}{4^2}.\frac{4.7}{5^2}...\frac{100.103}{101^2}\)
\(=\frac{1.4}{2^2}.\frac{2.5}{3^2}.\frac{3.6}{4^2}.\frac{4.7}{5^2}...\frac{98.101}{99^2}\frac{99.102}{100^2}\frac{100.103}{101^2}\)
\(=\frac{101.102.103}{1.2.3}\)
a) \(\left(-\frac{3}{4}\right)^2:\left(\frac{5}{4}\right)^2+14,7-1\frac{9}{25}\)
\(=\left[\left(-\frac{3}{4}\right):\frac{5}{4}\right]^2+\frac{147}{10}-\frac{34}{25}\)
\(=\left[\left(-\frac{3}{4}\right)\cdot\frac{4}{5}\right]^2+\frac{147}{10}-\frac{34}{25}\)
\(=\left(-\frac{3}{5}\right)^2+\frac{147}{10}-\frac{34}{25}=\frac{9}{25}+\frac{147}{10}-\frac{34}{25}=\left(\frac{9}{25}-\frac{34}{25}\right)+\frac{147}{10}=-1+\frac{147}{10}=\frac{137}{10}\)
b) \(\left(2\frac{1}{3}-1,5\right):\left(-6\frac{1}{6}+5\frac{1}{2}\right)+2,75\)
\(=\left(\frac{7}{3}-\frac{3}{2}\right):\left(-\frac{37}{6}+\frac{11}{2}\right)+\frac{11}{4}\)
\(=\frac{5}{6}:\left(-\frac{2}{3}\right)+\frac{11}{4}=\frac{5}{6}\cdot\left(-\frac{3}{2}\right)+\frac{11}{4}=-\frac{5}{4}+\frac{11}{4}=\frac{3}{2}\)
=1.1.2.2.3.3.....9.9/2.2.3.3.4.4....10.10
=1/10.10
=1/100
k to nha