chờ duyệt lâu quá
Tìm giá trị của các biểu thức sau
a)\(\frac{4^2\times4^3}{2^{10}}\)
b)\(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}\)
c)\(\frac{2^7\times9^3}{6^5\times8^2}\)
d)\(\frac{6^3+3\times6^2+3^3}{-13}\)
K
Khách
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8 tháng 8 2015
d)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(8+4+1\right)}{-13}=\frac{3^3\cdot13}{-13}=-27\)
giang làm a,b,c rồi nên làm d thôi
lười quá, hehe ^_^
8 tháng 8 2015
a) \(\frac{4^2\cdot4^3}{2^{10}}=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\frac{2^4.2^6}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)
b)\(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2.3\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2\right)^5.3^5}{\left(0,2\right)^6}=\frac{3^5}{0,2}=\frac{243.5}{1}=1215\)
c)\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3^5}{2^{11}.3^5}=\frac{1}{2^4}=\frac{1}{16}\)
7 tháng 8 2015
\(\frac{4^2\times4^3}{2^{10}}=\frac{\left(2^2\right)^2\times\left(2^2\right)^3}{2^{10}}=\frac{2^4\times2^6}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)
\(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2\times3\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2\right)^5\times3^5}{\left(0,2\right)^6}=\frac{3^5}{0,2}=\frac{243}{0,2}=243:\left(0,2\right)=243\times5=1215\)
\(\frac{2^7\times9^3}{6^5\times8^2}=\frac{2^7\times\left(3^2\right)^3}{\left(2\times3\right)^5\times\left(2^3\right)^2}=\frac{2^7\times3^6}{2^5\times3^5\times2^6}=\frac{3}{2^4}=\frac{3}{16}\)
\(\frac{6^3+3\times6^2+3^3}{-13}=\frac{\left(2\times3\right)^3+3\times6^2+3^3}{-13}=\frac{3\left(2^3\times3^2+6^2+3^2\right)}{-13}=\frac{3\times117}{-13}=\frac{351}{-13}=-27\)
1 tháng 8 2018
a. \(\frac{20^5.5^{10}}{100^5}\)
\(=\frac{20^5.\left(5^2\right)^5}{100^5}\)
\(=\frac{20^5.25^5}{100^5}\)
\(=\frac{500^5}{100^5}\)
\(=\left(\frac{500}{100}\right)^5\)
\(=5^5=3125\)
b. \(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\)
\(=\frac{\left(0,9\right)^5}{\left(0,3\right)^5.0,3}\)
\(=\left(\frac{0,9}{0,3}\right)^5.\frac{1}{0,3}\)
\(=3^5.\frac{1}{0,3}\)
\(=810\)
c. \(\frac{6^3+3.6^2+3^3}{-13}\)
\(=\frac{\left(3.2\right)^3+3.\left(3.2\right)^2+3^3}{-13}\)
\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}\)
\(=\frac{3^3.13}{-13}\)
\(=\left(-3\right)^3\)
\(=-27\)
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: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{2^3.(1+2+2^2)}}{{7^2}} = \frac{{2^3.7}}{{7^2}} = \frac{8}{7}\)
1 tháng 5 2015
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
1 tháng 10 2017
a) \(\frac{4^2.4^3}{2^{10}}=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\frac{2^4.2^6}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)
b) \(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2.3\right)^5}{\left(0,2\right)^6}=\frac{3^5.0,2^5}{0,2^6}=\frac{3^5}{0,2}=\frac{243}{0,2}=1215\)
c) \(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{2^5.3^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3}{2^{11}}=\frac{3}{2^4}=\frac{3}{16}\)
d) \(\frac{6^3+3.6^2+3^3}{-13}=\frac{6^2\left(6+3\right)+3^3}{-13}=\frac{6^2.9+3^2}{-13}=\frac{3^2\left(6^2+1\right)}{-13}=\frac{9.37}{-13}=\frac{333}{-13}\)
a, 1
b, 1215
c, \(\frac{3}{16}\)
d, \(\frac{3132}{-13}\)