$\frac{x}{\sqrt[]{x}-1}$- $\frac{2\sqrt[]{x}-x}{x-\sqrt[]{x}}$ =$\sqrt[]{x}$ -1
thay vào ta có
$\sqrt[]{2\sqrt[]{2}+3}$ -1 =$\sqrt[]{(\sqrt[]{1}+\sqrt[]{2}}$)²-1 =1+$\sqrt[]{2}$-1= $\sqrt[]{2}$
b)$\sqrt[]{4x+4}$ =$\sqrt[]{x+1}$+6
⇔2($\sqrt[]{x+1}$)= $\sqrt[]{x+1}$+6
⇔ $\sqrt[]{x+1}$=6
⇔x+1=36 ⇔x=35