Problema 26

$$\left( \frac{0,036}{0,2} \right)^2\cdot \left(\frac{0,0036}{0,04}\right)^{-2}=$$

a) 2

b) 4

c) $$2\cdot 10^{-10}$$

d) $$4\cdot 10^{-20}$$

e) $$4\cdot 10^{-10}$$


SOLUCIÓN:

Trasformando los decimales a fracción:

$$0,036=\frac{36}{1000} = \frac{9}{250}$$

$$0,2=\frac{2}{10} = \frac{1}{5}$$

$$0,0036=\frac{36}{10000} = \frac{9}{2500}$$

$$0,04=\frac{4}{100} = \frac{1}{25}$$


Entonces:

$$\left( \frac{0,036}{0,2} \right)^2\cdot \left(\frac{0,0036}{0,04}\right)^{-2}=\left( \frac{\frac{9}{250}}{\frac{1}{5}} \right)^2\cdot \left(\frac{\frac{9}{2500}}{\frac{1}{25}}\right)^{-2}=$$

$$=\left( \frac{9}{250}:\frac{1}{5} \right)^2\cdot \left(\frac{9}{2500}: \frac{1}{25}\right)^{-2}=$$

$$=\left( \frac{9}{250}\cdot \frac{5}{1} \right)^2\cdot \left(\frac{9}{2500}\cdot \frac{25}{1}\right)^{-2}= \left( \frac{9}{50} \right)^2\cdot \left(\frac{9}{100}\right)^{-2} $$


Recordando la propiedad de las potencias:

$$\left( \frac{a}{b}\right)^{-c} = \left( \frac{b}{a}\right)^{c}$$

y la propiedad de las potencias:

$$\left( a^c\cdot b^c \right) = \left( a\cdot b \right)^c$$

entonces se sigue:


$$\left( \frac{9}{50} \right)^2\cdot \left(\frac{9}{100}\right)^{-2}= \left( \frac{9}{50} \right)^2\cdot \left(\frac{100}{9}\right)^2 = \left( \frac{9}{50} \cdot \frac{100}{9}\right)^2= $$

$$\left( 2 \right)^2 = 4 $$

Recordando la proi