Đáp án+Giải thích các bước giải:
`A=\frac{-6}{\sqrt{x}+3}(x≥0)`
`\sqrt{x}+3≥3∀x≥0`
`⇔\frac{6}{\sqrt{x}+3}≤2`
`⇔\frac{-6}{\sqrt{x}+3}≥-2`
Dấu `"="` xảy ra khi `\sqrt{x}=0⇔x=0(tm)`
Vậy `A_{min}=-2⇔x=0`
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b,
`B=\frac{\sqrt{x}-5}{\sqrt{x}+2}(x≥0)`
`B=\frac{\sqrt{x}+2-7}{\sqrt{x}+2}`
`B=1+\frac{-7}{\sqrt{x}+2}`
`\sqrt{x}+2≥2∀x≥0`
`⇔\frac{7}{\sqrt{x}+2}≤\frac{7}{2}`
`⇔\frac{-7}{\sqrt{x}+2}≥-\frac{7}{2}`
`⇔1+\frac{-7}{\sqrt{x}+2}≥\frac{-5}{2}`
Dấu `"="` xảy ra khi `\sqrt{x}=0⇔x=0(tm)`
Vậy `B_{min}=\frac{-5}{2}⇔x=0`
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`C=\frac{x+\sqrt{x}+4}{\sqrt{x}+1}(x>0;x\ne4)`
`C=\sqrt{x}+\frac{4}{\sqrt{x}+1}`
`C=\sqrt{x}+1+\frac{4}{\sqrt{x}+1}-1`
Áp dụng bđt `Cauchy` cho `2` số dương ta có:
`\sqrt{x}+1+\frac{4}{\sqrt{x}+1}≥2\sqrt{(\sqrt{x}+1).\frac{4}{\sqrt{x}+1}}`
`⇔\sqrt{x}+1+\frac{4}{\sqrt{x}+1}≥4`
`⇔\sqrt{x}+1+\frac{4}{\sqrt{x}+1}-1≥3`
Dấu `"="` xảy ra khi
`\sqrt{x}+1=\frac{4}{\sqrt{x}+1}`
`⇔x+2\sqrt{x}+1=4`
`⇔x+2\sqrt{x}-3=0`
`⇔x-\sqrt{x}+3\sqrt{x}-3=0`
`⇔\sqrt{x}(\sqrt{x}-1)+3(\sqrt{x}-1)=0`
`⇔(\sqrt{x}+3)(\sqrt{x}-1)=0`
`\sqrt{x}+3≥3>0`
`⇒\sqrt{x}-1=0`
`⇔\sqrt{x}=1`
`⇔x=1(tm)`
Vậy `C_{min}=3⇔x=1`
