Đáp án+Giải thích các bước giải:
$1/$
Với `a>0`
Ta có:
`P=(1/\sqrta+\sqrta/(\sqrta+1)):\sqrta/(\sqrta+a)`
`\to P=(\sqrta+1+\sqrt{a}.\sqrt{a})/(\sqrta(\sqrta+1)):\sqrta/(\sqrta+a)`
`\to P=(\sqrta+1+a)/(\sqrta(\sqrta+1)):\sqrta/(\sqrta+a)`
`\to P=(a+\sqrta+1)/(a+\sqrta).(\sqrta+a)/\sqrta`
`\to P=(a+\sqrta+1)/(\sqrta)`
Vậy với `a>0` thì ` P=(a+\sqrta+1)/(\sqrta)`
$2/$
` P=(a+\sqrta+1)/(\sqrta)`
`\to P=13/3`
`⇔ (a+\sqrta+1)/(\sqrta)=13/3`
`⇔3(a+\sqrta+1)=13\sqrta`
`⇔3a+3\sqrta+3=13\sqrta`
`⇔3a-10\sqrta+3=0`
`⇔3a-9\sqrta-\sqrta+3=0`
`⇔3\sqrta(\sqrta-3)-(\sqrta-3)=0`
`⇔(3\sqrta-1)(\sqrta-3)=0`
\(⇔\left[ \begin{array}{l}3\sqrt a-1=0\\\sqrt a-3=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}3\sqrt a=1\\\sqrt a=3\end{array} \right.\)
\(⇔\left[ \begin{array}{l}\sqrt a=\dfrac{1}{3}\\\sqrt a=3\end{array} \right.\)
\(⇔\left[ \begin{array}{l}a=\dfrac{1}{9}(tm)\\a=9(tm)\end{array} \right.\)
Vậy `a∈\{1/9;3}` để `P=13/3`
