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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: \(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\cdot\dfrac{-7}{11}=-\dfrac{5}{11}\)
b: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\left(4+\dfrac{3}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{3}{2}\cdot5=\dfrac{15}{2}\)
c: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}=\dfrac{2}{15}\cdot\left(-4\right)+\dfrac{3}{15}=\dfrac{-8+3}{15}=\dfrac{-5}{15}=-\dfrac{1}{3}\)
d: \(=\dfrac{4}{9}\left(19+\dfrac{1}{3}-39-\dfrac{1}{3}\right)=\dfrac{4}{9}\cdot\left(-20\right)=-\dfrac{80}{9}\)
a-3; b-4; c-2; d-1
Giải thích:
\({3^7}{.3^3}=3^{7+3}=3^{10}\)
\({5^9}:{5^7}=5^{9-7}=5^2\)
\({2^{11}}:{2^8}=2^{11-8}=2^3\)
\({5^{12}}{.5^5}=5^{12+5}=5^{17}\)