a)x∈(9,49)
$\frac{\sqrt[]{x}+1 }{\sqrt[]{x}-3}$-2>0 (x>3)
⇔$\frac{\sqrt[]{x}+1 }{\sqrt[]{x}-3}$>2
⇔$\sqrt[]{x}$+1>2.($\sqrt[]{x}$-4)
⇔-$\sqrt[]{x}$>-9
⇔$\sqrt[]{x}$<9
⇔x<81
b)x≥0
$\frac{2.\sqrt[]{x}+1}{\sqrt[]{x}+2}$>- $\frac{1}{5}$
⇔6 $\sqrt[]{x}$+3>$\sqrt[]{x}$+2
⇔5$\sqrt[]{x}$>-1
⇔$\sqrt[]{x}$>- $\frac{1}{5}$
⇔x>0.04
