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GN
GV Nguyễn Trần Thành Đạt
Giáo viên
24 tháng 6 2023
\(\dfrac{8^{14}}{4^4.64^5}=\dfrac{\left(2^3\right)^{14}}{\left(2^2\right)^4.\left(2^5\right)^5}=\dfrac{2^{42}}{2^8.2^{25}}=2^{42-\left(8+25\right)}=2^9\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
24 tháng 6 2023
\(\dfrac{9^{10}.27^7}{81^7.3^{15}}=\dfrac{\left(3^2\right)^{10}.\left(3^3\right)^7}{\left(3^4\right)^7.3^{15}}=\dfrac{3^{20}.3^{21}}{3^{28}.3^{15}}=\dfrac{3^{20+21}}{3^{28+15}}=\dfrac{3^{41}}{3^{41}.3^2}=\dfrac{1}{3^2}=\dfrac{1}{9}\)
QD
1
19 tháng 9 2017
\(\frac{3^7.8^5}{6^6.\left(-2\right)^{12}}\)
\(=\frac{3^7.\left(2^3\right)^5}{2^6.3^6.2^{12}}\)
\(=\frac{3^7.2^{15}}{2^{18}.3^6}\)
\(=\frac{3}{2^3}=\frac{3}{8}\)
TT
20 tháng 12 2017
a,\(\dfrac{5}{6}+\left(-\dfrac{1}{2}\right)+\dfrac{3}{4}\)
\(=\dfrac{10}{12}+\left(-\dfrac{6}{12}\right)+\dfrac{9}{12}\)
\(=\dfrac{10-6+9}{12}=\dfrac{13}{12}\)
b,\(\left(0,75-\dfrac{1}{3}\right):\dfrac{7}{15}\)
\(=\left(\dfrac{3}{4}-\dfrac{1}{3}\right):\dfrac{7}{15}\)
\(=\left(\dfrac{9}{12}-\dfrac{4}{12}\right):\dfrac{7}{15}\)
\(=\dfrac{5}{12}:\dfrac{7}{15}\)
\(=\dfrac{25}{28}\)
c,\(\dfrac{7}{12}-\dfrac{3}{4}.\dfrac{5}{6}\)
\(=\dfrac{7}{12}-\dfrac{5}{8}\)
\(=\dfrac{14}{24}-\dfrac{15}{24}\)
\(=-\dfrac{1}{24}\)
d,\(\left(2\dfrac{1}{3}+1\dfrac{3}{4}\right).\dfrac{12}{13}\)
\(=\left(\dfrac{7}{3}+\dfrac{7}{4}\right).\dfrac{12}{13}\)
\(=\left(\dfrac{28}{12}+\dfrac{21}{12}\right).\dfrac{12}{13}\)
\(=\dfrac{49}{12}.\dfrac{12}{13}\)
\(=\dfrac{49}{13}\)
PN
20 tháng 12 2017
a) \(\dfrac{5}{6}+\left(\dfrac{-1}{2}\right)+\dfrac{3}{4}\)
\(=\dfrac{10}{12}-\dfrac{6}{12}+\dfrac{9}{12}\)
\(=\dfrac{13}{12}\)
b) \(\left(0,75-\dfrac{1}{3}\right):\dfrac{7}{15}\)
\(=\left(\dfrac{3}{4}-\dfrac{1}{3}\right).\dfrac{15}{7}\)
\(=\left(\dfrac{9}{12}-\dfrac{4}{12}\right).\dfrac{15}{7}\)
\(=\dfrac{5}{12}.\dfrac{15}{7}\)
\(=\dfrac{25}{28}\)
c) \(\dfrac{7}{12}-\dfrac{3}{4}.\dfrac{5}{6}\)
\(=\dfrac{7}{12}-\dfrac{5}{8}\)
\(=\dfrac{14}{24}-\dfrac{15}{24}\)
\(=\dfrac{-1}{24}\)
d) \(\left(2\dfrac{1}{3}+1\dfrac{3}{4}\right).\dfrac{12}{13}\)
\(=\left(\dfrac{7}{3}+\dfrac{7}{4}\right).\dfrac{12}{13}\)
\(=\left(\dfrac{28}{12}+\dfrac{21}{12}\right).\dfrac{12}{13}\)
\(=\dfrac{49}{12}.\dfrac{12}{13}\)
\(=\dfrac{49}{13}\)
AH
Akai Haruma
Giáo viên
27 tháng 9 2023
Lời giải:
1.
$(\frac{5}{6})^{10}.(\frac{3}{10})^{10}=(\frac{5}{6}.\frac{3}{10})^{10}=(\frac{1}{4})^{10}$
$=\frac{1}{4^{10}}$
2.
$(\frac{4}{7})^{19}: (\frac{-12}{35})^{19}=(\frac{4}{7}: \frac{-12}{35})^{19}=(\frac{-5}{3})^{19}$
3.
$(\frac{-3}{7})^7:\frac{-3}{5}=\frac{(-3)^7}{7^7}.\frac{5}{-3}=\frac{5.(-3)^6}{7^7}=\frac{5.3^6}{7^7}$
KK
21 tháng 12 2018
a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)=\dfrac{7}{8}.\left(\dfrac{10}{60}+\dfrac{24}{60}\right)=\dfrac{7}{8}.\dfrac{17}{30}=\dfrac{114}{240}\)
b)\(\dfrac{3}{2}-\dfrac{5}{6}\left(\dfrac{1}{2}\right)^2+\sqrt{4}=\dfrac{3}{2}-\dfrac{5}{6}.\dfrac{1}{4}+2=\dfrac{3}{2}-\dfrac{5}{24}+2=\dfrac{36}{24}-\dfrac{5}{24}+\dfrac{48}{24}=\dfrac{79}{24}\)c)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-1\dfrac{15}{17}+\dfrac{2}{3}=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-1\dfrac{15}{17}=1+\left(\dfrac{7}{21}+\dfrac{14}{21}\right)-\dfrac{32}{17}=1+1-\dfrac{32}{17}=2-\dfrac{32}{17}=\dfrac{34}{17}-\dfrac{32}{17}=\dfrac{2}{17}\)d)\(\left(-2\right)^3.\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)=-8.\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-8.\dfrac{2}{4}:\left(\dfrac{54}{24}-\dfrac{28}{24}\right)=-8.\dfrac{2}{4}:\dfrac{13}{12}=-4.\dfrac{12}{13}=\dfrac{-48}{13}\)e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)=16\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)+28\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)=\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right).\left(-\dfrac{5}{3}\right)=\left(\dfrac{120}{7}+\dfrac{196}{7}\right).\left(-\dfrac{5}{3}\right)=\dfrac{316}{7}.\left(-\dfrac{5}{3}\right)=-\dfrac{1580}{21}\)
Giải:
\(\dfrac{3^7.8^5}{6^6.\left(-2\right)^{12}}\)
\(=\dfrac{3^7.8^5}{6^6.2^{12}}\)
\(=\dfrac{3^7.2^{15}}{3^6.2^6.2^{12}}\)
\(=\dfrac{3^7.2^{15}}{3^6.2^{18}}\)
\(=\dfrac{3}{2^3}\)
\(=\dfrac{3}{8}\)
Vậy ...
\(\dfrac{3^7.8^5}{6^6.\left(-2\right)^{12}}=\dfrac{3^7.2^{15}}{2^6.3^6.2^{12}}=\dfrac{3}{8}\)