tính:
(\(\frac{9}{25}\)- 2 . 8) : (3\(\frac{4}{5}\)+ 0,2)
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Bài 1:
a) \(\left(\frac{9}{25}-2.18\right):\left(3\frac{4}{5}+0,2\right)\)
\(=\left(\frac{9}{25}-36\right):\left(\frac{19}{5}+\frac{1}{5}\right)\)
\(=\left(\frac{9}{25}-\frac{900}{25}\right):4\)
\(=-\frac{891}{25}.\frac{1}{4}\)
\(=-\frac{891}{100}\)
b) \(\frac{3}{8}.19\frac{1}{3}-\frac{3}{8}.33\frac{1}{3}\)
\(=\frac{3}{8}.\frac{58}{3}-\frac{3}{8}.\frac{100}{3}\)
\(=\frac{3}{8}\left(\frac{58}{3}-\frac{100}{3}\right)\)
\(=\frac{3}{8}\left(-\frac{42}{3}\right)\)
\(=\frac{3}{8}.\left(-14\right)\)
\(=-\frac{21}{4}\)
c) \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\frac{27}{23}+\frac{5}{21}-\frac{4}{23}+\frac{1}{2}+\frac{16}{21}\)
\(=\frac{27}{23}+\frac{5}{21}+\left(-\frac{4}{23}\right)+\frac{1}{2}+\frac{16}{21}\)
\(=\left[\frac{27}{23}+\left(-\frac{4}{23}\right)\right]+\left(\frac{5}{21}+\frac{16}{21}\right)+\frac{1}{2}\)
\(=1+1=2\)
d) \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
\(=\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{36}{45}\)
\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{36}{45}\right)\)
\(=1+1=2\)
a) \(=-\frac{91}{50}:4\)
\(=-\frac{91}{200}\)
b) \(=\frac{5}{18}-\frac{26}{5}+\frac{18}{5}\)
\(=-\frac{119}{90}\)
\(\left(\frac{9}{25}-2,18\right):\left(3\frac{4}{5}+0,2\right)\)
\(=\left(-1,82\right):4\)
\(=-0,455\)
Bài 1:
\(A=\frac{\frac{1}{12}-\frac{2}{9}-1}{\frac{5}{18}+\frac{-3}{4}-\frac{1}{9}}\)
\(A=\frac{\frac{1}{12}-\frac{2}{9}-\frac{18}{18}}{\frac{5}{18}+\frac{9}{12}-\frac{1}{9}}\)
\(A=\frac{1}{3}-\frac{18}{5}\)\(A=\frac{5}{15}-\frac{54}{15}\)
\(A=\frac{-49}{15}\)
a)
\(\frac{{{4^3}{{.9}^7}}}{{{{27}^5}{{.8}^2}}} = \frac{{{{\left( {{2^2}} \right)}^3}.{{\left( {{3^2}} \right)}^7}}}{{{{\left( {{3^3}} \right)}^5}.{{\left( {{2^3}} \right)}^2}}} =\frac{2^{2.3}.3^{2.7}}{3^{3.5}.2^{2.3}}= \frac{{{2^6}{{.3}^{14}}}}{{{3^{15}}{{.2}^6}}} = \frac{1}{3}\)
b)
\(\frac{{{{\left( { - 2} \right)}^3}.{{\left( { - 2} \right)}^7}}}{{{{3.4}^6}}} =\frac{(-2)^{3+7}}{3.(2^2)^6}= \frac{{{{\left( { - 2} \right)}^{10}}}}{{3.{{\left( {{2^{2.6}}} \right)}}}} = \frac{{{2^{10}}}}{{{{3.2}^{12}}}} = \frac{1}{{{{3.2}^2}}} = \frac{1}{{12}}\)
c)
\(\begin{array}{l}\frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,09} \right)}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} = \frac{{{{\left( {0,2} \right)}^5}.{{\left[ {{{\left( {0,3} \right)}^2}} \right]}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} = \frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,3} \right)}^6}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}}\\ = \frac{{{{\left( {0,3} \right)}^2}}}{{{{\left( {0,2} \right)}^2}}} = \frac{{0,9}}{{0,4}} = \frac{9}{4}\end{array}\)
d)
Cách 1: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{8 + 16 + 32}}{{49}} = \frac{{56}}{{49}} = \frac{8}{7}\)
Cách 2: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{2^3.(1+2+2^2)}}{{7^2}} = \frac{{2^3.7}}{{7^2}} = \frac{8}{7}\)
\(\left(\frac{9}{25}-2\cdot18\right):\left(3\frac{4}{5}+0,2\right)\)
\(=\left(0,36-36\right):\left(3,8+0,2\right)\)
\(=-35,64:4\)
\(=-8,91\)
(\(\frac{9}{25}\)- 2 . 8) : (3\(\frac{4}{5}\)+ 0,2)
= (\(\frac{9}{25}\)- 36) : (\(\frac{19}{5}\)+ \(\frac{1}{5}\))
= (\(\frac{9}{25}\)- \(\frac{36}{1}\)) : \(\frac{20}{5}\)
= (\(\frac{9}{25}\)- \(\frac{36.25}{1.25}\)) : \(\frac{20}{5}\)
= (\(\frac{9}{25}\)- \(\frac{900}{25}\)) : \(\frac{20}{5}\)
= \(\frac{-891}{25}\): \(\frac{1}{4}\)
= \(\frac{-891}{25}\). \(\frac{1}{4}\)
=\(\frac{-891}{100}\)