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a) \(A=\left\{x\in N|0\le x\le4\right\}\)
b) \(B=\left\{x\in N|x=4k;0\le k\le4;k\in N\right\}\)
c) \(C=\left\{x\in Z|x=\left(-3\right)^k;1\le k\le4;k\in N\right\}\)
d) \(D=\left\{x\in N|x=k^2;k=3a;1\le a\le4;a\in N\right\}\)
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1: \(\dfrac{16^{11}\cdot5^{40}}{10^{41}}=\dfrac{2^{44}\cdot5^{40}}{2^{41}\cdot5^{41}}=\dfrac{2^3}{5^1}=\dfrac{8}{5}\)
2: \(\dfrac{3^7\cdot8^5}{6^6\cdot\left(-2\right)^{12}}=\dfrac{3^7\cdot2^{15}}{2^6\cdot3^6\cdot2^{12}}=\dfrac{3}{2^3}=\dfrac{3}{8}\)
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Câu 1:
a) = \(\dfrac{-7}{2}\) x \(\dfrac{45}{32}\) = \(\dfrac{-315}{64}\)
b) = \(\dfrac{18}{7}\) : \(\dfrac{-27}{14}\) = \(\dfrac{18}{7}\) x \(\dfrac{14}{-27}\) = \(\dfrac{-4}{3}\)
c) = \(\dfrac{-3}{8}\) x ( \(\dfrac{5}{11}\) + \(\dfrac{6}{11}\) + 2 ) = \(\dfrac{-3}{8}\) x 3 = \(\dfrac{-9}{8}\)
Câu 2:
\(\dfrac{-3}{5}\) . x + \(\dfrac{7}{6}\) = \(\dfrac{5}{4}\)
\(\Leftrightarrow\) \(\dfrac{-3}{5}\) . x = \(\dfrac{5}{4}\) - \(\dfrac{7}{6}\)
\(\Leftrightarrow\) \(\dfrac{-3}{5}\) . x = \(\dfrac{1}{12}\)
\(\Leftrightarrow\) x = \(\dfrac{1}{12}\) : \(\dfrac{-3}{5}\)
\(\Leftrightarrow\) x = \(\dfrac{-5}{36}\)
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A = (1- 2) \(\times\) ( 4 - 3) \(\times\) (5 - 6) \(\times\) (8 - 7) \(\times\) (9 - 10) \(\times\) (12 - 11) \(\times\)(13 - 14)
A = (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1)
A = 1
a: \(=\dfrac{-2}{7}\cdot\dfrac{3}{2}=\dfrac{-3}{7}\)
b: \(=3\cdot\dfrac{7}{12}=\dfrac{7}{4}\)
c: \(=\dfrac{11}{12}\cdot\dfrac{16}{33}\cdot\dfrac{3}{5}=\dfrac{1}{3}\cdot\dfrac{4}{3}\cdot\dfrac{3}{5}=\dfrac{1}{3}\cdot\dfrac{4}{5}=\dfrac{4}{15}\)