The value of $\left(\frac{x}{y}\right)^{2a-3b}×\left(\frac{x}{y}\right)^{3b-4c}×\left(\frac{x}{y}\right)^{4c-2a}$
Answer & explanation
Correct answer: option 2
$\left(\frac{x}{y}\right)^{2a-3b+3b-4c+4c-2a}=\left(\frac{x}{y}\right)^0=1$
The value of $\left(\frac{x}{y}\right)^{2a-3b}×\left(\frac{x}{y}\right)^{3b-4c}×\left(\frac{x}{y}\right)^{4c-2a}$
Correct answer: option 2
$\left(\frac{x}{y}\right)^{2a-3b+3b-4c+4c-2a}=\left(\frac{x}{y}\right)^0=1$