Đáp án và giải thích các bước giải:
ĐKXĐ: `x≥0`
`A=1:({x+2\sqrt[x]-2}/{x\sqrt[x]+1}-{\sqrt[x]-1}/{x-\sqrt[x]+1}+1/{\sqrt[x]+1}`
`A=1:{x+2\sqrt[x]-2-(\sqrt[x]-1)(\sqrt[x]+1)+x-\sqrt[x]+1}/{(\sqrt[x]+1)(x-\sqrt[x]+1}`
`A=1:{x+\sqrt[x]}/{(\sqrt[x]+1)(x-\sqrt[x]+1}`
`A={(\sqrt[x]+1)(x-\sqrt[x]+1}/{x+\sqrt[x]}`
`A={x-\sqrt[x]+1}/{\sqrt[x]}`
`b)` Có : `x=4-2\sqrt[3]=(\sqrt[3])^2-2.\sqrt[3].1+1^2=(\sqrt[3]-1)^2`
`⇒\sqrt[x]=\sqrt[(\sqrt[3]-1)^2]=\sqrt[3]-1`
`⇒` `A={4-2\sqrt[3]-\sqrt[3]+1+1}/{\sqrt[3]-1}`
`A={6-3\sqrt[3]}/{\sqrt[3]-1}`
`A={(6-3\sqrt[3])(\sqrt[3]+1)}/{3-1}`
`A={3\sqrt[3]-3}/{2}`
`c)` Có : `A={x-\sqrt[x]+1}/{\sqrt[x]}`
`⇒` `A={\sqrt[x]-1}+1/\sqrt[x]`
Nhận xét :
`x≥0⇒\sqrt[x]≥0⇒1/{\sqrt[x]}>0`
`⇒\sqrt[x]+1+1/{\sqrt[x]}≥2\sqrt[\sqrt[x].1/{\sqrt[x]}]-1=2-1=1` (cô-si)
`⇒A_{min}=1⇔\sqrt[x]=1/{\sqrt[x]}⇔x=1(TMĐK)`
Vậy `A_{min}=1⇔x=1`
