Question

pls answer the attached question it's urgent​

Answer 1

Answer:

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Answer 2

★ To find :

Value of

$\left(\frac{1728}{729}\right)^{\frac{2}{3}$ × $\left(\frac{3^2}{\sqrt{144}}\right)^2$ × √5

★ Solution :

We know that,

$a^m \times a^n = a^{mn}$

$\left(\frac{1728}{729}\right)^{\frac{2}{3}$

$= \left(\left(\frac{1728}{729}\right)^2\right)^{\frac{1}{3}$

1728 = 12³ , 729 = 9³

$= \left(\left(\frac{12^3}{9^3}\right)^{2}\right)^{\frac{1}{3}$

$= \left(\frac{12^{3 \times 2}}{9^{3 \times 2}}\right)^{\frac{1}{3}}$

$=\left(\frac{12^6}{9^6}\right)^{\frac{1}{3}$

$\bold{= \frac{12^2}{9^2}}$

$\left(\frac{3^2}{\sqrt{144}}\right)^2 =$

144 = 12²

$= (\frac{3^2}{12})^2$

$\bold{= \frac{3^4}{12^2}}$

by putting the simplified versions together .

9 = 3²

$\frac{\cancel{12^2}}{(3^2)^2} \times \frac{3^4}{\cancel{12^2}} \times \sqrt{5}$

$\leadsto \frac{1}{3^4} \times \cancel{3^4} \times \sqrt{5}$

$\purple{\leadsto = \sqrt{5}}$