a.
Ta có: `f(x) = x/(x+1)`
`\to f(1) = 1/(1+1) = 1/2`
`\to f(f(1)) = (1/2)/(1/2+1) = 1/3`
`\to f(f(f(1))) = (1/3)/(1/3+1) = 1/4`
Vậy `f(f(f(1))) = 1/4`
b.
Ta có: `f(\sqrt{x}) = 5/6`
`⇔\sqrt{x}/(\sqrt{x}+1) = 5/6`
`⇔6\sqrt{x} = 5(\sqrt{x}+1)`
`⇔6\sqrt{x} = 5\sqrt{x} + 5`
`⇔\sqrt{x} = 5`
`⇔x = 25`
Vậy `x = 25` khi `f(\sqrt{x}) = 5/6`