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The value of $\left(\frac{x}{y}\right)^{2a-3b}×\left(\frac{x}{y}\right)^{3b-4c}×\left(\frac{x}{y}\right)^{4c-2a}$ |
$\frac{x}{y}$ 1 0 $\left(\frac{x}{y}\right)^{2a+3b+4c}$ |
1 |
$\left(\frac{x}{y}\right)^{2a-3b+3b-4c+4c-2a}=\left(\frac{x}{y}\right)^0=1$ |
