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e: \(=\dfrac{5^{30}\cdot3^{20}}{3^{15}\cdot5^{30}}=3^5=243\)
\(A=\left(-\dfrac{5}{7}+\dfrac{8}{5}\right):\dfrac{91}{8}+\left(-\dfrac{2}{7}-\dfrac{3}{5}\right):\dfrac{91}{8}\)
\(=\dfrac{31}{35}:\dfrac{91}{8}+\dfrac{-31}{35}:\dfrac{91}{8}\)
\(=\dfrac{248}{3185}+\dfrac{-248}{3185}\)
= 0
\(B=\dfrac{13}{15}:\left(\dfrac{4}{5}-\dfrac{3}{7}\right)+\dfrac{13}{15}:\left(\dfrac{2}{5}-\dfrac{1}{9}\right)\)
\(=\dfrac{13}{15}:\dfrac{13}{35}+\dfrac{13}{15}:\dfrac{13}{45}\)
\(=\dfrac{7}{3}+3\)
\(=\dfrac{16}{3}\)
\(a,13\dfrac{3}{5}-\left(8\dfrac{3}{5}-4\dfrac{3}{4}\right)\)
\(=\dfrac{68}{5}-\dfrac{43}{5}+\dfrac{19}{4}\)
\(=5+\dfrac{19}{4}\)
\(=\dfrac{20}{4}+\dfrac{19}{4}=\dfrac{39}{4}\)
\(------\)
\(b,11\dfrac{1}{4}-\left(2\dfrac{5}{7}+5\dfrac{1}{4}\right)\)
\(=\dfrac{45}{4}-\dfrac{19}{7}-\dfrac{21}{4}\)
\(=\left(\dfrac{45}{4}-\dfrac{21}{4}\right)-\dfrac{19}{7}\)
\(=6-\dfrac{19}{7}\)
\(=\dfrac{42}{7}-\dfrac{19}{7}=\dfrac{23}{7}\)
\(\left(-\dfrac{4}{3}+\dfrac{5}{13}\right):\dfrac{2}{7}-\left(\dfrac{9}{4}+\dfrac{8}{13}\right):\dfrac{2}{7}\\ =\left(-\dfrac{4}{3}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right):\dfrac{2}{7}\\ =-\dfrac{595}{156}:\dfrac{2}{7}\\ =-\dfrac{595}{156}.\dfrac{7}{2}=-\dfrac{4165}{312}\)
\(A=\dfrac{\left(17+\dfrac{1}{4}-4-\dfrac{3}{16}-13-\dfrac{5}{6}\right)\cdot\left(-\dfrac{4}{7}\right)+\dfrac{27}{4}}{\left(5+\dfrac{2}{7}-5-\dfrac{1}{3}\right):\left(6+\dfrac{2}{3}-4-\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{37}{84}+\dfrac{27}{4}}{-\dfrac{1}{21}:\dfrac{13}{6}}=\dfrac{-1963}{6}\)
Bài 1:
a) \(3^7:3^5-\left(\dfrac{5}{17}\right)^0=3^{7-5}-1=3^2-1=9-1=8\)
b) \(\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{1}{2}+2\right)^3\)
\(=\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{5}{2}\right)^3\)
\(=\left(\dfrac{5}{2}\right)^{10}\)
c) \(8.\left(\dfrac{1}{4}\right)^3+\left(\dfrac{2}{27}\right)^0-\dfrac{1}{8}\)
\(=8.\dfrac{1}{64}+1-\dfrac{1}{8}\)
\(=\dfrac{1}{8}+1-\dfrac{1}{8}\)
\(=1\)
Bài 2:
a) \(\dfrac{3^4.4^4}{6^4}=\dfrac{3^4.\left(2^2\right)^4}{\left(2.3\right)^4}=\dfrac{3^4.2^8}{2^4.3^4}=\dfrac{2^8}{2^4}=2^4=16\)
b) \(\dfrac{15^3}{10^3}=\dfrac{\left(3.5\right)^3}{ \left(2.5\right)^3}=\dfrac{3^3.5^3}{2^3.5^3}=3^3:2^3=\dfrac{27}{8}\)
c) \(\dfrac{4^2.12^5}{9^2.2^{10}}=\dfrac{\left(2^2\right)^2.\left[3.\left(2^2\right)\right]^5}{\left(3^2\right)^2.2^{10}}=\dfrac{2^4.3^5.2^{10}}{3^4.2^{10}}=2^4.3=16.3=48\)
d) \(\dfrac{6^2+5.2^2+4}{15}=\dfrac{\left(2.3\right)^2+5.2^2+2^2}{15}=\dfrac{2^2.3^2+5.2^2+2^2}{15}=\dfrac{2^2\left(3^2+5+1\right)}{15}=\dfrac{2^2.15}{15}=2^2=4\)
Bài 3:
a) \(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(\dfrac{-5}{12}\right)^2}\)
\(=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.-1}{\left[\dfrac{2}{5}.\left(\dfrac{-5}{12}\right)\right]^2}\)
\(=\dfrac{\left(\dfrac{2}{3}\right)^3. \left(\dfrac{-3}{4}\right)^2.-1}{\left(\dfrac{-1}{6}\right)^2}\)
\(=\left(\dfrac{2}{3}\right)^3.\left[\left(\dfrac{-3}{4}\right).-6\right]^2.-1\)
\(=\left(\dfrac{2}{3}\right)^3.\left(\dfrac{9}{2}\right)^2.-1\)
\(=\left(\dfrac{2}{3}\right)^2.\dfrac{2}{3}.\left(\dfrac{9}{2}\right)^2.-1\)
\(=\left(\dfrac{2}{3}.\dfrac{9}{2}\right)^2.\dfrac{2}{3}.-1\)
\(=9.\dfrac{2}{3}.-1\)
\(=6.-1=-6\)
b) \(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{\left(2.3\right)^6+\left(2.3\right)^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^6+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}=\dfrac{3^6.73}{-73}=\dfrac{3^6}{-1}=\left(-3\right)^6\)
\(#Wendy.Dang\)
Lần sau bnn gửi từng bài thôi nha, chứ như vầy nhiều quá thì làm không nổi mất. đánh máy nãy giờ lú luôn gòi nè :))
( 2x -1)2 - \(\dfrac{1}{4}\) = 2 ( -2\(\dfrac{1}{5}\) + \(\dfrac{2}{7}\) - \(\dfrac{12}{13}\)) - ( - \(\dfrac{5}{7}\)+\(\dfrac{1}{13}\))
(2x -1)2 = 2 + \(\dfrac{1}{4}\) = -\(\dfrac{11}{5}\) + \(\dfrac{2}{7}\) - \(\dfrac{12}{13}\) + \(\dfrac{5}{7}\) - \(\dfrac{1}{13}\)
(2x -1) 2 = \(\dfrac{9}{4}\) = - \(\dfrac{11}{5}\) + ( \(\dfrac{2}{7}+\dfrac{5}{7}\)) - ( \(\dfrac{12}{13}+\dfrac{1}{13}\))
\(\left[{}\begin{matrix}2x-1=\dfrac{3}{2}\\2x-1=-\dfrac{3}{2}\end{matrix}\right.\) = - \(\dfrac{11}{5}\) + 1 - 1
\(\left[{}\begin{matrix}2x=\dfrac{3}{2}+1\\2x=-\dfrac{3}{2}+1\end{matrix}\right.\) = -11/5
\(\left[{}\begin{matrix}2x=2,5\\2x=-0,5\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=1,25\\x=-0,25\end{matrix}\right.\)
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
\(E=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{5751}{25}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
\(=\dfrac{\dfrac{25}{108}.\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\left(\dfrac{-44}{21}\right)}\)
\(=\dfrac{53,25+\dfrac{187}{4}}{\dfrac{-25}{11}}\)
\(=\dfrac{100}{\dfrac{-25}{11}}\)
\(=-44\)
\(=\left(-\dfrac{3}{4}+\dfrac{5}{13}\right).\dfrac{7}{2}-\left(\dfrac{9}{4}+\dfrac{8}{13}\right).\dfrac{7}{2}\\ =\dfrac{7}{2}\left(-\dfrac{3}{4}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right)\\ =\dfrac{7}{2}\left(-\dfrac{12}{4}-\dfrac{3}{13}\right)\\ =\dfrac{7}{2}\left(-3-\dfrac{3}{13}\right)\\ =\dfrac{7}{2}\left(-\dfrac{39}{13}-\dfrac{3}{13}\right)\\ =\dfrac{7}{2}\left(-\dfrac{42}{13}\right)\\ =\dfrac{7.\left(-42\right)}{2.13}\\ =-\dfrac{147}{13}\)