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a) \(\dfrac{6^2.6^3}{3^5}=\dfrac{6^5}{3^5}=2^5\)
b) \(\dfrac{25^2.4^2}{5^5.\left(-2\right)^5}=\dfrac{5^4.2^4}{5^5.\left(-2\right)^5}=\dfrac{1.1}{5.\left(-2\right)}=\dfrac{1}{-10}=\dfrac{-1}{10}\)
( Mình giải được có 2 bài thôi. ) :)
a) \(\dfrac{6^2.6^3}{3^5}=\dfrac{6^5}{3^5}=3^5\)
b)\(\dfrac{25^2.4^2}{5^5.\left(-2\right)^5}=\dfrac{\left(25.4\right)^2}{\left(5.\left(-2\right)\right)^5}=\dfrac{\left(100\right)^2}{\left(-10\right)^5}=\dfrac{1}{-10}\)
Còn phần c mình cx đang tắc
b) \(\left(-2\dfrac{3}{4}+\dfrac{1}{2}\right)^2\)
\(=\left(\dfrac{-11}{4}+\dfrac{1}{2}\right)^2\)
\(=\left(\dfrac{-11}{4}+\dfrac{2}{4}\right)^2\)
\(=\left(\dfrac{-9}{4}\right)^2\)
\(=\dfrac{81}{16}\)
\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)
\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)
\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)
\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)
\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)
\(=7-\dfrac{26}{5}\)
\(=\dfrac{9}{5}\)
\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)
\(=\dfrac{2}{3}+\dfrac{21}{8}\)
\(=\dfrac{79}{24}\)
\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)
\(=\dfrac{31}{4}:\dfrac{49}{8}\)
\(=\dfrac{62}{49}\)
\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)
\(\dfrac{\left(0,125\right)^5.\left(2,4\right)^5}{\left(-0,3\right)^5.\left(0.01\right)^3}=\dfrac{\left(0,125\right)^5.\left(-0,3\right)^5.\left(-8\right)^5}{\left(-0,3\right)^5.\left(0.01\right)^3}=\dfrac{\left(0,125\right)^5.\left(-8\right)^5}{\left(0.01\right)^3}=\dfrac{\left(-8.0,125\right)^5}{\left(0.01\right)^3}=\dfrac{-1}{0.000001}=-1000000\)
\(=\dfrac{3^5}{-\left(0.003\right)^3\cdot0.09}=-10^{11}\)
\(P=\left(0,5-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right):\left(-2\right)\)
\(=\left(-\dfrac{1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right).\left(-\dfrac{1}{2}\right)\)
\(=\left(\dfrac{-5-6}{10}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{12}\)
\(=-\dfrac{11}{10}:\left(-3\right)+\dfrac{1}{4}\)
\(=-\dfrac{11}{10}.\left(-\dfrac{1}{3}\right)+\dfrac{1}{4}=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{37}{60}\)
Vậy \(P=\dfrac{37}{60}\)
\(Q=\left(\dfrac{2}{25}-1,008\right):\dfrac{4}{7}:\left[\left(3\dfrac{1}{4}-6\dfrac{5}{9}\right):2\dfrac{2}{17}\right]\)
\(=\left(\dfrac{2}{25}-\dfrac{126}{125}\right):\dfrac{4}{7}:\left[\left(\dfrac{13}{4}-\dfrac{59}{9}\right).\dfrac{36}{17}\right]\)
\(=-\dfrac{116}{125}.\dfrac{7}{4}:\left(-\dfrac{119}{36}.\dfrac{36}{17}\right)\)
\(=\dfrac{-29.7}{125}:\left(-7\right)=\dfrac{29}{125}\)
Vậy \(Q=\dfrac{29}{125}\)
Có \(\left(-2\dfrac{3}{4}+\dfrac{1}{2}\right)^2\)=\(\left(\dfrac{-5}{4}+\dfrac{2}{4}\right)^2\)=\(\left(\dfrac{-3}{4}\right)^2\)=\(\dfrac{\left(-3\right)^2}{4^2}=\dfrac{9}{16}\)
Có \(\dfrac{\left(0,125\right)^5.\left(2,4\right)^5}{\left(-0,3\right)^5.\left(0,01\right)^3}=\dfrac{\left(0,125.2,4\right)^5}{\left(-0,3\right)^5.\left(0,01\right)^3}=\dfrac{\left(0,3\right)^5}{\left(-0.3\right)^5.\left(0,01\right)^3}=\dfrac{1}{-1.\left(0,01\right)^3}=\dfrac{1}{-\left(0,01\right)^3}\)