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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) \(\frac{{38}}{7} + \left( { - 3,25} \right) - \frac{{17}}{7} + 4,55\)
\(\begin{array}{l} = \left( {\frac{{38}}{7} - \frac{{17}}{7}} \right) + \left( {4,55 - 3,25} \right)\\ = \frac{{38 - 17}}{7} + 1,3 = \frac{{21}}{7} +1,3\\ = 3 + 1,3 = 4,3\end{array}\)
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a) \(\frac{{ - 3}}{7}.\frac{2}{5} + \frac{2}{5}.\left( { - \frac{5}{{14}}} \right) - \frac{{18}}{{35}}\)
\(\begin{array}{l} = \frac{2}{5}.\left( {\frac{{ - 3}}{7} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\left( {\frac{{ - 6}}{{14}} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\frac{{ - 11}}{{14}} - \frac{{18}}{{35}} = \frac{{ - 11}}{{35}} - \frac{{18}}{{35}} = \frac{{ -29}}{{35}}\end{array}\)
b) \(\left( {\frac{2}{3} - \frac{5}{{11}} + \frac{1}{4}} \right):\left( {1 + \frac{5}{{12}} - \frac{7}{{11}}} \right)\)
\(\begin{array}{l} = \left( {\frac{{2.11.4}}{{3.11.4}} - \frac{{5.3.4}}{{11.3.4}} + \frac{{1.3.11}}{{4.3.11}}} \right):\left( {\frac{11.12}{11.12} + \frac{{5.11}}{{12.11}} - \frac{{7.12}}{{11.12}}} \right)\\ = \left( {\frac{{88 - 60 + 33}}{{121}}} \right):\left( { \frac{{121+55 - 84}}{{121}}} \right)\\ = \frac{{61}}{{121}}:\frac{{92}}{{121}} = \frac{{61}}{{121}}.\frac{{121}}{{92}}= \frac{{61}}{{92}}\end{array}\)
c) \(\left( {13,6 - 37,8} \right).\left( { - 3,2} \right)\)
\( = \left( { - 24,2} \right).\left( { - 3,2} \right) = 77,44\)
d) \(\left( { - 25,4} \right).\left( {18,5 + 43,6 - 16,8} \right):12,7\)
\(\begin{array}{l} = \left( { - 25,4} \right).\left( {62,1 - 16,8} \right):12,7\\ = \left( { - 25,4} \right).45,3:12,7\\ = \left( { - 25,4} \right):12,7.45,3\\ = (- 2).45,3 = - 90,6\end{array}\)
a: \(=\dfrac{2}{5}\cdot\left(-\dfrac{3}{7}-\dfrac{5}{14}\right)-\dfrac{18}{35}\)
\(=\dfrac{2}{5}\cdot\dfrac{-6-5}{14}-\dfrac{18}{35}\)
\(=\dfrac{2}{5}\cdot\dfrac{-11}{14}-\dfrac{18}{35}=-\dfrac{22}{70}-\dfrac{18}{35}=\dfrac{-58}{70}=-\dfrac{29}{35}\)
b: \(=\dfrac{88-60+33}{132}:\dfrac{132+55-84}{132}\)
\(=\dfrac{61}{132}\cdot\dfrac{132}{103}=\dfrac{61}{103}\)
c: \(=-24.2\cdot\left(-3.2\right)=24.2\cdot3.2=77.44\)
d: \(=\dfrac{-25.4}{12.7}\cdot45.3=-2\cdot45.3=-90.6\)
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A = (4+\(\frac{1}{5}\)) . \(\frac{18}{19}\)+ (2+\(\frac{8}{5}\)) . \(\frac{21}{5}\)
A= \(\frac{21}{5}\).18/19 + 18/5 . 21/5
A= 21/5 (18/19 + 18/5)
A= 21/5 . 432/95
A= 9288/95
b= 25/2. (3+2/7) - 23/7. (5 + 1/2)
b= 25/2 . 23/7 - 23/7 . 11/2
b= 23/7 (25/2 -11/2)
b=23/7 . 7
b= 23
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P/s: Bài này chỉ tính được giá trị "gần đúng" của biểu thức thôi nhé!
\(\frac{21}{54}+\frac{3}{75}:\frac{\left(\frac{39}{65}+0,415-\frac{33}{600}\right)\frac{21}{91}}{7^2-18,25+13\frac{15}{36}-16\frac{17}{102}}\)
\(\Leftrightarrow\frac{21}{54}+\frac{3}{75}:\frac{\left(0,6+0,415-0,055\right)0,23}{49-18,25+\frac{483}{36}-\frac{1649}{102}}\)
\(\Leftrightarrow\frac{21}{54}+\frac{3}{75}:\frac{\left(0,6+0,415-0,055\right)0,23}{49-18,25+13,41-16,1}\)
\(\Leftrightarrow\frac{21}{54}+\frac{3}{75}:\frac{0,96.0,23}{28,06}\Leftrightarrow\frac{21}{54}+\frac{3}{75}:\frac{0,2208}{28,06}\)
\(\Leftrightarrow\frac{21}{54}+\left(\frac{3}{75}:\frac{0,2208}{28,06}\right)\Leftrightarrow\frac{21}{54}+\left(\frac{3}{75}.\frac{28,06}{0,2208}\right)=\frac{21}{54}+\frac{61}{12}=\frac{197}{36}\)
P/s: Giải bài này mà mệt cả đầu =((
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\(=8\frac{4}{17}-2\frac{5}{9}+3\frac{4}{17}=\left(8+3\right)\frac{4}{17}-2\frac{5}{9}=11\frac{4}{17}-2\frac{5}{9}\)
\(\frac{19.5^{22}-5^{13}-25^{18}}{\left(7.5^{17}\right)^2}\) \(=\frac{19.5^{22}-5^{13}-5^{36}}{7^2.5^{34}}\) \(=\frac{5^{13}\left(19.5^9-1-5^{23}\right)}{7^2.5^{34}}\)
\(=\frac{19.5^9-1-5^{23}}{7^2.5^{21}}\)
Hình như đề bài có vấn đề bn ak
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