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\(=\dfrac{25}{9}.\dfrac{3}{10}+\dfrac{25}{9}.\dfrac{7}{10}+\dfrac{25}{3}=\dfrac{25}{9}\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\dfrac{25}{3}=\dfrac{25}{9}+\dfrac{25}{3}=\dfrac{100}{9}\)
\(E=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{5751}{25}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
\(=\dfrac{\dfrac{25}{108}.\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\left(\dfrac{-44}{21}\right)}\)
\(=\dfrac{53,25+\dfrac{187}{4}}{\dfrac{-25}{11}}\)
\(=\dfrac{100}{\dfrac{-25}{11}}\)
\(=-44\)
a, \(\dfrac{3}{7}\)\(x\)- \(\dfrac{2}{3}\)\(x\) = \(\dfrac{10}{21}\)
(\(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)) \(\times\) \(x\) = \(\dfrac{10}{21}\)
- \(\dfrac{5}{21}\) \(\times\) \(x\) = \(\dfrac{10}{21}\)
\(x\) = \(\dfrac{10}{21}\) : (-\(\dfrac{5}{21}\))
\(x\) = -2
b, \(\dfrac{7}{35}\) : (\(x-\dfrac{1}{3}\)) = - \(\dfrac{2}{25}\)
\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{7}{35}\) : (- \(\dfrac{2}{25}\))
\(x\) - \(\dfrac{1}{3}\) = - \(\dfrac{5}{2}\)
\(x\) = - \(\dfrac{5}{2}\) + \(\dfrac{1}{3}\)
\(x\) = - \(\dfrac{13}{6}\)
c, 3.(\(x\) - \(\dfrac{1}{2}\)) - 5.(\(x\) + \(\dfrac{3}{5}\)) = - \(x\)+ \(\dfrac{1}{5}\)
3\(x\) - \(\dfrac{3}{2}\) - 5\(x\) - 3 = - \(x\) + \(\dfrac{1}{5}\)
- \(x\) + 5\(x\) - 3\(x\) = - \(\dfrac{3}{2}\) - 3 - \(\dfrac{1}{5}\)
\(x\) = - \(\dfrac{47}{10}\)
\(a,\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow x\left(\dfrac{3}{7}-\dfrac{2}{3}\right)=\dfrac{10}{21}\\ \Rightarrow x.-\dfrac{5}{21}=\dfrac{10}{21}\\ \Rightarrow x=-2\\ b,\dfrac{7}{35}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow\dfrac{1}{5}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow x-\dfrac{1}{3}=-\dfrac{5}{2}\\ \Rightarrow x=-\dfrac{13}{6}\\ c,3.\left(x-\dfrac{1}{2}\right)-5.\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\\ \Rightarrow3x-\dfrac{3}{2}-5x+5=-x+\dfrac{1}{5}\)
\(\Rightarrow x\left(3-5\right)-\dfrac{3}{2}+5=-x+\dfrac{1}{5}\\ \Rightarrow-2x-\dfrac{13}{2}=-x+\dfrac{1}{5}\\ \Rightarrow-x-\dfrac{13}{5}=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{1}{5}-\dfrac{13}{5}\\ \Rightarrow x=-\dfrac{12}{5}.\)
\(=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right)\cdot230.04+46.75}{\left(\dfrac{13}{10}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
\(=\dfrac{53.25+46.75}{\dfrac{139}{30}:\dfrac{-41}{21}}=100:\left(\dfrac{139}{30}\cdot\dfrac{-21}{41}\right)\)
\(=100\cdot\dfrac{-400}{973}=\dfrac{-40000}{973}\)
\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)
\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)
Bài 2:
a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)
\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)
\(=-\dfrac{3}{5}\)
b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)
\(\Leftrightarrow8x-1=5\)
\(\Leftrightarrow8x=6\)
hay \(x=\dfrac{3}{4}\)
\(A=\dfrac{25^{10}+1}{25^{10}-1}=\dfrac{25^{10}-1+2}{25^{10}-1}=\dfrac{25^{10}-1}{25^{10}-1}+\dfrac{2}{25^{10}-1}=1+\dfrac{2}{25^{10}-1}\left(1\right)\)
\(B=\dfrac{25^{10}-1}{25^{10}-3}=\dfrac{25^{10}-3+2}{25^{10}-3}=\dfrac{25^{10}-3}{25^{10}-3}+\dfrac{2}{25^{10}-3}=1+\dfrac{2}{25^{10}-3}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A< B\)
Ta có : \(\dfrac{25^{10}+1}{25^{10}-1}=\dfrac{25^{10}-1+2}{25^{10}-1}=\dfrac{25^{10}-1}{25^{10}-1}+\dfrac{2}{25^{10}-1}\)\(=1+\dfrac{2}{25^{10}-1}\)
Ta có : \(\dfrac{25^{10}-1}{25^{10}-3}=\dfrac{25^{10}-3+2}{25^{10}-3}=\dfrac{25^{10}-3}{25^{10}-3}+\dfrac{2}{25^{10}-3}=1+\dfrac{2}{25^{10}-3}\)
Vì \(25^{10}-1>25^{10}-3\)
\(\Rightarrow\dfrac{2}{25^{10}-1}< \dfrac{2}{25^{10}-3}\)
\(\Rightarrow1+\dfrac{2}{25^{10}-1}< 1+\dfrac{2}{25^{10}-3}\)
\(\Rightarrow A< B\)
Vậy \(A< B\)