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ND
1
VM
20 tháng 9 2019
b) \(\frac{2^9.4^{10}}{8^8}=\frac{2^9.\left(2^2\right)^{10}}{\left(2^3\right)^8}=\frac{2^9.2^{20}}{2^{24}}=\frac{2^{29}}{2^{24}}=2^5=32.\)
Chúc bạn học tốt!
24 tháng 10 2019
\(\left(\frac{3}{7}-\frac{5}{2}-\frac{3}{5}\right)-\left(-\frac{4}{7}-\frac{3}{2}+\frac{2}{5}\right)=\frac{3}{7}-\frac{5}{2}-\frac{3}{5}+\frac{4}{7}+\frac{3}{2}-\frac{2}{5}\)
\(=\left(\frac{3}{7}+\frac{4}{7}\right)+\left(-\frac{5}{2}+\frac{3}{2}\right)+\left(-\frac{3}{5}-\frac{2}{5}\right)\)
\(=1+\left(-1\right)+\left(-1\right)\)
\(=-1\)
Hok tốt nha^^
3 tháng 8 2017
\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}-\frac{5}{10}+\frac{5}{11}+\frac{5}{12}}+\frac{\frac{3}{2}+1+\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}+\frac{5}{4}}\)
\(=\frac{3.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}+\frac{3.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}{5.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{3}{5}\)
\(=\frac{6}{5}\)
24 tháng 6 2018
\(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{2}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+\frac{10}{4}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=10\)
24 tháng 6 2018
\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{2}{8}+\frac{1}{9}\)
Tách 9=1+1+...+1 ( có 9 số 1)
\(\Rightarrow A=1+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{2}{8}+1\right)+\left(\frac{1}{9}+1\right)\)
\(A=\frac{10}{10}+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{8}+\frac{10}{9}\)
\(A=10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
\(\Rightarrow A:B=\frac{10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}=10\) ( vì \(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\ne0\) )
Vậy \(A:B=10\)
HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023
a)
\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)
b)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)
XO
26 tháng 10 2019
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
XO
26 tháng 10 2019
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
giup voi
a) \(\frac{8^5.27^3}{6^9.4^4}=\frac{\left(2^3\right)^5.\left(3^3\right)^3}{2^9.3^9.\left(2^2\right)^4}=\frac{2^{15}.3^9}{2^9.3^9.2^8}=\frac{2^{15}.3^9}{2^{17}.3^9}=\frac{1}{2^2}=\frac{1}{4}\)
b) \(\left(\frac{-2}{3}\right)^3:\frac{8}{3}-\left|-\frac{5}{3}\right|=\frac{-8}{27}:\frac{8}{3}-\frac{5}{3}\)
\(=-\frac{1}{9}-\frac{5}{3}=-\frac{1}{9}-\frac{15}{9}=-\frac{16}{9}\)