Đáp án:
$\frac{4\sqrt[]{x}}{3(x-\sqrt[]{x}+1)}$
=$\frac{4}{3(\sqrt[]{x}-1+\frac{1}{\sqrt[]{x}})}$
ta có $\sqrt[]{x}$ +$\frac{1}{\sqrt[]{x}}$$\geq$ 2
=> $\frac{4}{3(\sqrt[]{x}-1+\frac{1}{\sqrt[]{x}})}$ $\leq$ $\frac{4}{3}$
dấu "=" xảy ra <=> x=1
Vậy Max = $\frac{4}{3}$
$\frac{4\sqrt[]{x}}{3(x-\sqrt[]{x}+1)}$ ≥0
dáu "=" xảy ra <=> x=0
Min = 0