Đáp án: $P_{min}=4⇔x=4$
Giải thích các bước giải:
`c)A=\frac{x+\sqrt{x}}{(\sqrt{x}-1)^2};B=\frac{\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-1)}`
`⇒P=A÷B=\frac{x+\sqrt{x}}{(\sqrt{x}-1)^2}÷\frac{\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-1)}`
`=\frac{\sqrt{x}(\sqrt{x}+1)\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)^2(\sqrt{x}-1)}`
`=\frac{x}{\sqrt{x}-1}`
Với $x>1$ thì:
`\frac{1}{P}=\frac{\sqrt{x}-1}{x}=\frac{-1}{x}+\frac{1}{\sqrt{x}}`
`=\frac{1}{4}-(\frac{1}{x}-\frac{1}{\sqrt{x}}+\frac{1}{4})=\frac{1}{4}-(\frac{1}{\sqrt{x}}-\frac{1}{2})^2`
Do `(\frac{1}{\sqrt{x}}-\frac{1}{2})^2≥0`
`⇒\frac{1}{P}=\frac{1}{4}-(\frac{1}{\sqrt{x}}-\frac{1}{2})^2≤\frac{1}{4}`
$⇒P≥4$
Dấu bằng xảy ra `⇔(\frac{1}{\sqrt{x}}-\frac{1}{2})^2=0`
`⇔\frac{1}{\sqrt{x}}-\frac{1}{2}=0⇔\frac{1}{\sqrt{x}}=\frac{1}{2}`
`⇔\sqrt{x}=2⇔x=4(tm)`
