a) 2ⁿ.4= 128

=> 2ⁿ=128:4

=>2ⁿ=32

Mà: 2⁵=32

=> x=5

a) \(4\frac{7}{10}< 6\frac{7}{10}\)(4 < 6)

b) \(3\frac{4}{15}< 3\frac{11}{15}\)(4/15 < 11/15)

c) \(5\frac{1}{9}>2\frac{2}{5}\)(5 > 2)

d) \(2\frac{2}{3}=2\frac{10}{15}\)(10/15 = 2/3)

a)    \(\frac{14}{21}=\frac{2}{3}=\frac{4}{6}\)

       \(\frac{60}{72}=\frac{5}{6}\)

Vì   \(\frac{4}{6}< \frac{5}{6}\)

nên       \(\frac{4}{21}< \frac{60}{72}\)

4 7/10 < 6 7/10

3 4/15 <3 11/15

5 1/9 > 2 2/5

2 2/5 > 2 10/15

ko btttttttttttt

a) 54⁴=3⁴.9⁴.2⁴=3⁴.(3²)⁴.2⁴=3⁴.3⁸.2⁴=3¹².2⁴

21¹²=3¹²7¹² > 3¹².2⁴=54⁴

b)10²⁰=10^2.10=100¹⁰ > 90¹⁰

c)5³⁹=\(\frac{5^{40}}{5}=\frac{5^{2.20}}{5}=\frac{25^{20}}{5}\) < \(\frac{27^{20}}{5}< 27.27^{20}=27^{21}\)

\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)

\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot...\cdot\frac{80}{81}\cdot\frac{99}{100}\)

\(B=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot...\cdot\frac{8.10}{9.9}\cdot\frac{9.11}{10.10}\)

\(B=\frac{\left(1\cdot2\cdot...\cdot8\cdot9\right).\left(3\cdot4\cdot...\cdot10\cdot11\right)}{\left(2\cdot3\cdot..\cdot9\cdot10\right).\left(2\cdot3\cdot...\cdot9\cdot10\right)}\)

\(B=\frac{1\cdot2\cdot...\cdot8\cdot9}{2\cdot3\cdot...\cdot9\cdot10}\cdot\frac{3\cdot4\cdot...\cdot10\cdot11}{2\cdot3\cdot...\cdot9\cdot10}\)

\(B=\frac{1}{10}\cdot\frac{11}{2}=\frac{11}{20}\)

Vì 20 < 21 nên 11/20 > 11/21

Vậy ..... 

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