a/ ĐK: $x\ne 0;y\ne 0$

$\dfrac{24x^4y^7}{16x^2y^9}\\=\dfrac{8x^2y^7.3x^2}{8x^2y^7.2y^2}\\=\dfrac{3x^2}{2y^2}$

Vậy $\dfrac{24x^4y^7}{16x^2y^9}=\dfrac{3x^2}{2y^2}$ với $x\ne 0;x\ne 0$

b/ ĐK: $y\ne 0;x\ne y$

$\dfrac{5x(x-y)}{7y(y-x)}\\=\dfrac{5x(x-y)}{-7y(x-y)}\\=-\dfrac{5x}{7y}$

Vậy $\dfrac{5x(x-y)}{7y(y-x)}=-\dfrac{5x}{7y}$ với $y\ne 0;x\ne y$

c/ ĐK: $x\ne \pm 2$

$\dfrac{x^2-4x+4}{x^2-4}\\=\dfrac{(x-2)^2}{(x-2)(x+2)}\\=\dfrac{x-2}{x+2}$

Vậy $\dfrac{x^2-4x+4}{x^2-4}=\dfrac{x-2}{x+2}$ với $x\ne \pm 2$

$\\$

`a,`

`(24x^4y^7)/(16x^2y^9)`

`= (3x^2)/(2y^2)`

`b,`

`(5x(x-y))/(7y(y-x))`

`= (5x(x-y))/(-7y(x-y))`

`= (-5x)/(7y)`

`c.`

`(x^2-4x+4)/(x^2-4)`

`=(x-2)^2/((x-2)(x+2))`

`= (x-2)/(x+2)`