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a)
\({5^7}{.5^5} = 5^{7+5}={5^{12}}\)
\({9^5} :{8^0} = {9^5}:1 = {9^5}\)
\(2^{10}:64.16 = 2^{10}:2^6.2^4 = 2^{10-6+4} = 2^8\)
b)
\(\begin{array}{l}54297 = 5.10000 + 4.1000 + 2.100 + 9.10 + 7\\ = {5.10^4} + {4.10^3} + {2.10^2} + 9.10 + 7\end{array}\)
\(\begin{array}{l}2023 = 2.1000 +0.100+2.10 + 3\\ = {2.10^3}+ 2.10 +3\end{array}\)
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
7. \(H_2SO_4\): Axit sunfuric
\(Fe\left(OH\right)_2\) : Sắt(II) hiđroxit
\(NaCl\) : Natri clorua
\(P_2O_5\): Điphotphopenta oxit
8. \(FeCl_3\): Sắt (III) clorua
\(Fe\left(OH\right)_3\): Sắt(III) hiđroxit
\(HCl\): Axit clohiđric
\(SO_3\): Lưu huỳnh trioxit
a-3; b-4; c-2; d-1
Giải thích:
\({3^7}{.3^3}=3^{7+3}=3^{10}\)
\({5^9}:{5^7}=5^{9-7}=5^2\)
\({2^{11}}:{2^8}=2^{11-8}=2^3\)
\({5^{12}}{.5^5}=5^{12+5}=5^{17}\)