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6 tháng 11 2017
a, \(\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
=\(\dfrac{2\cdot\left(2^3\right)^4\cdot\left(3^3\right)^2+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot\left(3^2\right)^4}\)
=\(\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
=\(\dfrac{2^{11}\cdot3^6\cdot\left(2^2+3^3\right)}{2^{10}\cdot3^7\cdot\left(2^4+5\cdot3\right)}\)
=\(\dfrac{2^{11}\cdot3^6\cdot31}{2^{10}\cdot3^7\cdot31}\)
=\(\dfrac{2}{3}\)
6 tháng 11 2017
b, \(\dfrac{\dfrac{8}{27}\cdot\dfrac{9}{16}\cdot\left(-1\right)}{\dfrac{4}{25}\cdot\dfrac{-125}{1728}}\)
=\(\dfrac{\dfrac{8\cdot9\cdot\left(-1\right)}{27\cdot16}}{\dfrac{4\cdot\left(-125\right)}{25\cdot1728}}\)
=\(\dfrac{\dfrac{-1}{6}}{\dfrac{-5}{432}}\)
=\(\dfrac{-1}{6}\cdot\dfrac{-432}{5}\)
=\(\dfrac{72}{5}\)
31 tháng 5 2015
\(A=\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)
\(B=\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=2\)
\(C=\frac{4^5\cdot9^{4-2\cdot6^9}}{2^{10}\cdot3^8+6^8\cdot20}=0\)
15 tháng 8 2016
A=\(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)
19 tháng 3 2020
a) Ta có: \(\left(-2\right)^3+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)
\(=-8+\frac{1}{2}\cdot8-5+64\)
\(=-8+4-5+64=55\)
b) Ta có: \(\left(\frac{-3}{4}+\frac{2}{7}\right):\frac{2}{3}+\left(\frac{-1}{4}+\frac{5}{7}\right):\frac{2}{3}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}\right)\cdot\frac{3}{2}+\left(\frac{-1}{4}+\frac{5}{7}\right)\cdot\frac{3}{2}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{5}{7}\right)\cdot\frac{3}{2}\)
\(=0\cdot\frac{3}{2}=0\)
c) Ta có: \(\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\frac{2^{10}\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot20}=\frac{2\left(2^9\cdot9^4-6^9\right)}{6^8\left(2^2+20\right)}=\frac{-1}{3}\)
19 tháng 3 2020
a) ( -2 )3 + \(\frac{1}{2}:\frac{1}{8}\) - √25 + \(|-64|\)
= \(\frac{-8}{1}\) + \(\frac{1}{2}.\frac{8}{1}\) - \(\frac{5}{1}\) + \(\frac{64}{1}\)
= \(\frac{-16}{2}+\frac{1}{2}.\frac{8}{1}-\frac{10}{2}+\frac{128}{2}\)
= \(\frac{-16}{2}+\frac{8}{2}-\frac{10}{2}+\frac{128}{2}\)
= \(\frac{-16+8-10+128}{2}\) = \(\frac{110}{2}\) = 55
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
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: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{2^3.(1+2+2^2)}}{{7^2}} = \frac{{2^3.7}}{{7^2}} = \frac{8}{7}\)
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