a) Ta có:
$f(x)=\dfrac{x}{x +1}$
$f(1)=\dfrac{1}{1 + 1}=\dfrac12$
$f(f(1))=f\left(\dfrac12\right)=\dfrac{\dfrac12}{\dfrac12 + 1}=\dfrac13$
$f(f(f(1)))=f\left(\dfrac13\right)=\dfrac{\dfrac13}{\dfrac13 +1}=\dfrac14$
b) $f(\sqrt x)=\dfrac56$
$\to \dfrac{\sqrt x}{\sqrt x +1} =\dfrac56$
$\to 6\sqrt x = 5(\sqrt x +1)$
$\to \sqrt x = 5$
$\to x = 25$