Question

If ${\left(\frac{-2}{7}\right)}^{-4}\times {\left(\frac{-5}{7}\right)}^2$ is $\frac{1225}{m}$, then the value of $m$ is

Answer 1

${\left(\frac{-2}{7}\right)}^{-4}\times {\left(\frac{-5}{7}\right)}^2={\left(\frac{7}{-2}\right)}^4\times {\left(\frac{-5}{7}\right)}^2$

$={\left(\frac{-7}{2}\right)}^4\times {\left(\frac{-5}{7}\right)}^2$ ...$[\because \left(\frac{7}{-2}\right)=\left(\frac{-7}{2}\right)]$

$=\frac{(-7{)}^4}{2^4}\times \frac{(-5{)}^2}{7^2}$

$=\frac{7^4\times (-5{)}^2}{2^4\times 7^2}$ ...$[\because (-7{)}^4=7^4]$

$=\frac{7^2\times (-5{)}^2}{2^4}$

$=\frac{49\times (-5)\times (-5)}{16}$

$=\frac{1225}{16}$