`a)B=(1/(sqrtx-3)+4/(x-9))(sqrtx-3)`
`=((sqrtx+3)/(x-9)+4/(x-9))(sqrtx-3)`
`=(sqrtx+7)/(x-9)*(sqrtx-3)`
`=(sqrtx+7)/(sqrtx+3)`
`b)P=A:B`
`sqrtx/(sqrtx+3):(sqrtx+7)/(sqrtx+3)`
`=sqrtx/(sqrtx+7)`
Vì `x>=0=>sqrtx>=0`
`x>=0=>sqrtx+7>=7>0`
`=>P>=0`
Dấu "=" xảy ra khi `x=0`
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a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
Ta có : \(B=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\sqrt{x-3}\)
\(=1+\dfrac{4}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3+4}{\sqrt{x}+3}=\dfrac{\sqrt{x}+7}{\sqrt{x}+3}\)
b, Ta có : \(P=\dfrac{\dfrac{\sqrt{x}}{\sqrt{x}+3}}{\dfrac{\sqrt{x}+7}{\sqrt{x}+3}}=\dfrac{\sqrt{x}}{\sqrt{x}+7}\)
- Thấy : \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}+7>0\end{matrix}\right.\)
\(\Rightarrow P\ge0\)
Vậy \(MinP=0\Leftrightarrow x=0\)
