`ĐKXĐ:x>0`
`a,P=({1}/{\sqrt{x}}+{\sqrt{x}}/{\sqrt{x}+1}):{\sqrt{x}}/{x+\sqrt{x}}`
`=[{\sqrt{x}+1}/{\sqrt{x}(\sqrt{x}+1)}+{x}/{\sqrt{x}(\sqrt{x}+1)}]:{\sqrt{x}}/{\sqrt{x}(\sqrt{x}+1)}`
`={\sqrt{x}+1+x}/{\sqrt{x}(\sqrt{x}+1)}:{1}/{\sqrt{x}+1}`
`={x+\sqrt{x}+1}/{\sqrt{x}(\sqrt{x}+1)}.(\sqrt{x}+1)`
`={x+\sqrt{x}+1}/{\sqrt{x}}`
`={x}/{\sqrt{x}}+{\sqrt{x}}/{\sqrt{x}}+{1}/{\sqrt{x}}`
`=\sqrt{x}+1+{1}/{\sqrt{x}}`
Vậy với `x>0` thì `P=\sqrt{x}+1+{1}/{\sqrt{x}}`
`b,x=4(TM)`
Thay `x=4` vào `P` có:
`P=\sqrt{4}+1+{1}/{\sqrt{4}}=2+1+1/2=3+1/2=7/2`
Vậy với `x=4` thì `P=7/2`
`c,P=13/3`
`⇔\sqrt{x}+1+{1}/{\sqrt{x}}=13/3`
`⇔\sqrt{x}+1+{1}/{\sqrt{x}}-13/3=0`
`⇔{3x}/{3\sqrt{x}}+{3\sqrt{x}}/{3\sqrt{x}}+{3}/{3\sqrt{x}}-{13\sqrt{x}}/{3\sqrt{x}}=0`
`⇒3x+3\sqrt{x}+3-13\sqrt{x}=0`
`⇔3x-10\sqrt{x}+3=0`
`⇔3x-9\sqrt{x}-\sqrt{x}+3=0`
`⇔3\sqrt{x}(\sqrt{x}-3)-(\sqrt{x}-3)=0`
`⇔(\sqrt{x}-3)(3\sqrt{x}-1)=0`
`⇔` \(\left[ \begin{array}{l}\sqrt{x}-3=0\\3\sqrt{x}-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}\sqrt{x}=3\\3\sqrt{x}=1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=9\\\sqrt{x}=\dfrac{1}{3}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=9(TM)\\x=\dfrac{1}{9}(TM)\end{array} \right.\)
Vậy với `x\in{9;1/9}` thì `P=13/3`
