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3/7.9/11+3/7.5/11-3/7.25/11
=3/7.(9/11+5/11-25/11)
=3/7.(-1)
=-3/7
\(\frac{1}{4}C=\frac{9}{16\cdot25}+\frac{3}{25\cdot28}+\frac{6}{28\cdot34}+\frac{6}{34\cdot40}\)
\(\frac{1}{4}C=\frac{1}{16}-\frac{1}{25}+\frac{1}{25}-\frac{1}{28}+\frac{1}{28}-\frac{1}{34}+\frac{1}{34}-\frac{1}{40}\)
\(\frac{1}{4}C=\frac{1}{16}-\frac{1}{40}\)
\(C=\frac{3}{20}\)
\(=\dfrac{3^{28}\cdot2^{21}\cdot5^{10}}{2^9\cdot3^3\cdot2^2\cdot3\cdot5^{12}}=\dfrac{3^{28}}{3^4}\cdot\dfrac{2^{21}}{2^{11}}\cdot\dfrac{1}{5^2}=\dfrac{3^{24}\cdot2^{10}}{5^2}\)
a, \(\frac{7.25-49}{7.24+21}=\frac{7.25-7.7}{7.24+7.3}=\frac{7.\left(25-7\right)}{7.\left(24+3\right)}=\frac{7.18}{7.27}=\frac{7.9.2}{7.9.3}=\frac{2}{3}\)
b, \(\frac{2.\left(-13\right).9.10}{-3.4.\left(-5\right).26}=\frac{2.13.\left(-3\right).3.\left(-2\right).\left(-5\right)}{-3.\left(-2\right).\left(-2\right).\left(-5\right).2.13}=\frac{-3}{2}\)
\(\frac{1}{3}\cdot\frac{7.24+3}{7.25-4}=\frac{1}{3}\cdot\frac{7.24+3}{7.24+7-4}=\frac{1}{3}\cdot\frac{7\cdot24+3}{7.23+3}=\frac{1}{3}.1=\frac{1}{3}\)
Câu 1:
a: \(16^4=2^{16}< 4^{16}\)
c: \(3^{25}=9\cdot3^{23}>8\cdot3^{23}\)
d: \(2^{50}=32^{10}>25^{10}=5^{20}\)