`C1:`
Theo bđt `Cô-si`:
`x+1≥2\sqrtx`
`⇔x+\sqrtx+1≥3\sqrtx`
`⇔C=\sqrtx/(x+\sqrtx+1)≤\sqrtx/(3\sqrtx)=1/3`
Dấu `=` xảy ra `⇔x+1=2\sqrtx⇒x=1`
Vậy $Max_C=\dfrac{1}{3}⇔x=1$
`C2`:
`C=\sqrtx/(x+\sqrtx+1)`
`C=[1/3(x+\sqrtx+1)-1/3(x-2\sqrtx+1)]/(x+\sqrtx+1)`
`C=1/3-1/3. (\sqrtx-1)^2/(x+\sqrtx+1)≤1/3`
Dấu `=` xảy ra `⇔1/3. (\sqrtx-1)^2/(x+\sqrtx+1)=0⇒x=1`
Vậy `Max_C=1/3⇔x=1`