$\frac{x\sqrt{x}+4x+\sqrt{x}}{(\sqrt{x}-2)^2}$ tìm GTNN
2 câu trả lời
Đáp án:
`A_{min}=0⇔x=0`
Giải thích các bước giải:
`ĐK` : `x≥0,x\ne4`
Đặt `A={x\sqrt[x]+4x+\sqrt[x]}/{(\sqrt[x]-2)^2}`
`A={\sqrt[x](x+4\sqrt[x]+1)}/{(\sqrt[x]-2)^2}`
Nhận xét :
`x≥0⇒\sqrt[x]≥0⇒4\sqrt[x]≥0`
`⇒` `x+4\sqrt[x]+1≥1`
`⇒` `x+4\sqrt[x]+1>0`
Mà : `(\sqrt[x]-2)^2>0`
`⇒` `{\sqrt[x](x+4\sqrt[x]+1)}/{(\sqrt[x]-2)^2}≥0`
`⇔` `A≥0`
`⇒` `A_{min}=0`
`⇔` `{\sqrt[x](x+4\sqrt[x]+1)}/{(\sqrt[x]-2)^2}=0`
`⇒` `\sqrt[x](x+4\sqrt[x]+1)=0`
`⇔` \(\left[ \begin{array}{l}\sqrt{x}=0\\x+4\sqrt{x}+1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x+4\sqrt{x}=-1\end{array} \right.\)
Mà : `x+4\sqrt[x]≥0`
`⇒` `x=0` `(TMĐK)`
Vậy `A_{min}=0⇔x=0`
Bạn tham khảo nhé.
`\frac{x\sqrt{x}+4x+\sqrt{x}}{(\sqrt{x}-2)^2}(x\ne4,x>=0)`
`=\frac{\sqrt{x}(x+4\sqrt{x}+1)}{(\sqrt{x}-2)^2}`
Vì `x>=0`
`=>\sqrt{x}>=0`
`=>x+4\sqrt{x}+1>=1>0`
Mà `(\sqrt{x}-2)^2>0`
`=>\frac{\sqrt{x}(x+4\sqrt{x}+1)}{(\sqrt{x}-2)^2}>=0`
Dấu `=` xảy ra khi `\frac{\sqrt{x}(x+4\sqrt{x}+1)}{(\sqrt{x}-2)^2}=0`
`<=>\sqrt{x}(x+4\sqrt{x}+1)=0`
`\text{TH1}:` `\sqrt{x}=0`
`<=>x=0(tm)`
`\text{TH2}:` `x+4\sqrt{x}+1=0`
`<=>x+4\sqrt{x}=-1(\text{Vô lý})`
Vậy `GTNNNN=0` khi `x=0`