(-0,2)2 . 5 \(-\)\(\frac{8^2.9^4}{3^7.4^3}\)
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F
1
PN
3
15 tháng 7 2019
\(\frac{9^3\cdot2^{10}\cdot27^5}{4^5\cdot81^6}=\frac{3^6\cdot2^{10}\cdot3^{15}}{2^{10}\cdot3^{24}}=\frac{1}{3^3}=\frac{1}{27}\)
\(\frac{27^4\cdot2^5-3^{11}\cdot4^3}{8^2\cdot9^6}=\frac{3^{12}\cdot2^5}{2^6\cdot3^{12}}-\frac{3^{11}\cdot2^6}{2^6\cdot3^{12}}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)
=))
XO
15 tháng 7 2019
a) \(\frac{9^3.2^{10}.27^5}{4^5.81^6}\)= \(\frac{\left(3^2\right)^3.2^{10}.\left(3^3\right)^5}{\left(2^2\right)^5.\left(3^4\right)^5}\)= \(\frac{3^{2.3}.2^{10}.3^{3.5}}{2^{2.5}.3^{4.5}}\)= \(\frac{3^6.2^{10}.3^{15}}{2^{10}.3^{20}}\)= \(\frac{3^{21}.2^{10}}{2^{10}.3^{20}}\)= \(\frac{3^{20}.2^{10}.3}{2^{10}.3^{20}}\)= \(3\)
TA
4 tháng 9 2020
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)
4 tháng 9 2020
Bài giải
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)
17 tháng 8 2015
1) \(7.4^x=7.4^3\Leftrightarrow4^x=4^3;x=3\)
2) \(\frac{3}{2.5^x}=\frac{3}{2.5^{12}}\Leftrightarrow5^x=5^{12};x=12\)
\(2^x=2.2^8=2^9;x=9\)
4) \(5.3^x=7.3^5-2.3^5\Leftrightarrow5.3^x=3^5.\left(7-2\right)\)
\(\Leftrightarrow3^5.x=3^5.5;x=5\)
MN
26 tháng 3 2021
a) 32 . 53 + 92 = 9 . 125 + 81
= 1 125 + 81 = 1 206
b) 83 : 42 - 52 = 512 : 16 - 25 = 32 - 25 = 7
c) 33 . 92 - 52.9 + 18 : 6 = 27 . 81 - 25 . 9 + 3
= 2 187 - 225 + 3 = 1 962 + 3 = 1 965
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: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{2^3.(1+2+2^2)}}{{7^2}} = \frac{{2^3.7}}{{7^2}} = \frac{8}{7}\)
VM
11 tháng 8 2019
Tính:
2) \(\left(\frac{2}{3}\right)^3-\left(\frac{3}{4}\right)^2.\left(-1\right)^5\)
\(=\frac{8}{27}-\frac{9}{16}.\left(-1\right)\)
\(=\frac{8}{27}-\left(-\frac{9}{16}\right)\)
\(=\frac{371}{432}.\)
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