Question

$[\sqrt{\frac{{15}^2}{{27}^2}}-\sqrt{\frac{5^2}{{12}^2}}]$ $\times \sqrt{\frac{9^2}{4^2}}$.

Answer 1

$[\sqrt{\frac{{15}^2}{{27}^2}}-\sqrt{\frac{5^2}{{12}^2}}]\sqrt{\frac{9^2}{4^2}}$

$=(\frac{15}{27}-\frac{5}{12})\left(\frac{9}{4}\right)=\frac{60-45}{108}\times \frac{9}{4}=\frac{25}{12}\times \frac{1}{4}=\frac{5}{16}$