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i) \(\frac{25^9}{5^{16}}-5^3:5\)
\(=\frac{\left(5^2\right)^9}{5^{16}}-5^2\)
\(=\frac{5^{18}}{5^{16}}-25\)
\(=5^2-25\)
\(=25-25\)
\(=0.\)
k) \(\frac{5}{7}-\left|\frac{2}{-7}\right|\)
\(=\frac{5}{7}-\left|\frac{-2}{7}\right|\)
\(=\frac{5}{7}-\frac{2}{7}\)
\(=\frac{3}{7}.\)
l) \(\frac{3^6.3^4}{9^3}\)
\(=\frac{3^{6+4}}{\left(3^2\right)^3}\)
\(=\frac{3^{10}}{3^6}\)
\(=3^4.\)
\(=81.\)
Chúc bạn học tốt!
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a, 5\(\dfrac{4}{27}\) + \(\dfrac{6}{23}\) + 0,25 - \(\dfrac{4}{27}\) + \(\dfrac{17}{23}\)
= 5 + (\(\dfrac{4}{27}\) - \(\dfrac{4}{27}\)) + (\(\dfrac{6}{23}\) + \(\dfrac{17}{23}\)) + 0,25
= 5 + 1 + 0,25
= 6,25
b, 16.(\(\dfrac{1}{2}\))3 - \(\dfrac{3}{5}\): 0,75
= 16.\(\dfrac{1}{8}\) - 0,8
= 2 - 0,8
= 1,2
a)\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}>1024^9\)
b) \(9^{12}=\left(3^2\right)^{12}=3^{24}\) và \(27^7=\left(3^3\right)^7=3^{21}\)
=> \(9^{12}>27^7\)
a, 1024 mũ 9 = 2 mũ 10 .9 = 2 mũ 90 < 2 mũ 100
b, 27 mũ 7 = 3 mũ 3.7 =3 mũ 21 < 3 mũ 24 = 3 mũ 2.12 = 9 mũ 12
c,2 mũ 161 > 2 mũ 160 = 2 mũ 4.40 = 16 mũ 40 > 13 mũ 40
1, \(\frac{5}{21}+0,5-\frac{19}{23}+\frac{16}{21}-\frac{4}{23}\)
\(=\frac{5+16}{21}-\frac{19+4}{23}+0,5\)
\(=1-1+0,5\)
\(=0,5\)
2,\(\frac{27}{23}+\frac{5}{21}-\frac{4}{23}+\frac{6}{21}+\frac{1}{2}\)
\(=\frac{27-4}{23}+\frac{5+6}{21}+\frac{1}{2}\)
\(=1+\frac{11}{21}+\frac{1}{2}\)
\(=\frac{42+22+21}{42}\)
\(=\frac{85}{42}\)
Bài 1:
a, Ta có: \(\dfrac{27}{23}+\dfrac{5}{21}-\dfrac{4}{23}+\dfrac{16}{21}+\dfrac{1}{2}\)
= \(\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
=\(\dfrac{23}{23}+\dfrac{21}{21}+\dfrac{1}{2}\) = \(1+1+\dfrac{1}{2}\)
= \(2\dfrac{1}{2}\)
