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T
4 tháng 10 2023
\(a,\dfrac{15^3}{5^4}\)
\(=\dfrac{\left(3\cdot5\right)^3}{5^4}\)
\(=\dfrac{3^3\cdot5^3}{5^4}\)
\(=\dfrac{3^3}{5}\)
\(=\dfrac{27}{5}\)
\(---\)
\(b,\dfrac{21^3}{7^4}\)
\(=\dfrac{\left(3\cdot7\right)^3}{7^4}\)
\(=\dfrac{3^3\cdot7^3}{7^4}\)
\(=\dfrac{3^3}{7}\)
\(=\dfrac{27}{7}\)
\(---\)
\(c,\dfrac{6^6}{3^8}\)
\(=\dfrac{\left(2\cdot3\right)^6}{3^8}\)
\(=\dfrac{2^6\cdot3^6}{3^8}\)
\(=\dfrac{2^6}{3^2}\)
\(=\dfrac{64}{9}\)
#\(Toru\)
Y
6 tháng 6 2023
\(\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{23}}\times\dfrac{\dfrac{1}{3}-0,25-0,2}{1\dfrac{1}{6}-0,875-0,7}\)
\(=\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{23}-\dfrac{1}{23}}\times\dfrac{\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}-\dfrac{7}{10}}\)
\(=\dfrac{\dfrac{1}{3} +\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{1}{3}\times2+\dfrac{1}{7}\times2-\dfrac{1}{23}\times2}\times\dfrac{\dfrac{2}{6}-\dfrac{2}{8}-\dfrac{2}{10}}{7\times\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}\)
\(=\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{2\times\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}\right)}\times\dfrac{2\times\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}{7\times\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}\)
\(=\dfrac{1}{2}\times\dfrac{2}{7}\)
\(=\dfrac{1}{7}\)
VV
25 tháng 10 2021
\(1\)/
\(a\)) \(=\left(\dfrac{7}{5}-\dfrac{8}{7}\right)+\dfrac{17}{5}:0,6\)
\(=\dfrac{9}{35}+\dfrac{17}{3}\)
\(=\dfrac{622}{105}\)
\(b\)) \(=\dfrac{11}{6}+\dfrac{-14}{15}\)
\(=\dfrac{9}{10}\)
\(c\)/ \(=\dfrac{7}{4}-\dfrac{2}{3}\)
\(=\dfrac{13}{12}\)
16 tháng 8 2021
Ta có: \(D=\left(2\dfrac{2}{15}\cdot\dfrac{9}{17}\cdot\dfrac{3}{32}\right):\left(-\dfrac{3}{17}\right)\)
\(=\dfrac{32}{15}\cdot\dfrac{9}{17}\cdot\dfrac{3}{32}\cdot\dfrac{-17}{3}\)
\(=\dfrac{-9}{15}=-\dfrac{3}{5}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
30 tháng 9
bài1
a) \(\dfrac{7}{6}-\dfrac{13}{12}+\dfrac{3}{4}\)
=\(\dfrac{14}{12}-\dfrac{13}{12}+\dfrac{9}{12}\)
=\(\dfrac{1}{12}+\dfrac{9}{12}\)
=\(\dfrac{10}{12}=\dfrac{5}{6}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
30 tháng 9
bài 1
b)\(1\dfrac{1}{2}.(\dfrac{-4}{5})\) + \(\dfrac{3}{10}\)
= \(\dfrac{3}{2}.\left(-\dfrac{4}{5}\right)+\dfrac{3}{10}\)
= \(-\dfrac{6}{5}+\dfrac{3}{10}\)
=\(-\dfrac{12}{10}+\dfrac{3}{10}\)
=\(-\dfrac{9}{10}\)
N
14 tháng 9 2021
\(a.\dfrac{-42}{12}+\dfrac{9}{12}-\dfrac{17}{12}=-\dfrac{25}{6}\)
\(b.-\dfrac{1}{12}-\dfrac{5}{4}+\dfrac{1}{3}=-\dfrac{1}{12}-\dfrac{15}{12}+\dfrac{4}{12}=-1\)
J
14 tháng 9 2021
a) \(\dfrac{-7}{2}+\dfrac{3}{4}-\dfrac{17}{12}=\dfrac{-42}{12}+\dfrac{9}{12}-\dfrac{17}{12}=\dfrac{-42+9-17}{12}=-\dfrac{50}{12}=-\dfrac{25}{6}\)
b)\(-\dfrac{1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)=-\dfrac{1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)=-\dfrac{2}{24}-\left(\dfrac{63}{24}-\dfrac{8}{24}\right)\)
\(=-\dfrac{2}{24}-\dfrac{63}{24}+\dfrac{8}{24}=-\dfrac{57}{24}=-\dfrac{19}{8}\)
LP
15 tháng 12 2021
\(a.=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}=\dfrac{1+3-5}{2}-\dfrac{2+5-7}{3}=\dfrac{-1}{2}\)
\(b.\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}=\left(-\dfrac{5}{12}\right)^2:\dfrac{5}{12}=\dfrac{5}{12}\)
\(=\dfrac{10}{15}+\dfrac{-5}{15}+\dfrac{7}{15}=\dfrac{12}{15}=\dfrac{4}{5}\)
\(\dfrac{1}{3}+\dfrac{7}{15}=\dfrac{5}{15}+\dfrac{7}{15}=\dfrac{12}{15}=\dfrac{4}{5}\)