Đáp án:
`2)` `x=1`
`3)` `x=4`
`4)` `x\in ∅`
Giải thích các bước giải:
`2)` `P={4x}/{\sqrt{x}-3}\quad (x> 0;x\ne 9)`
Để `P={\sqrt{x}+3}/{\sqrt{x}-3}`
`=>{4x}/{\sqrt{x}-3}={\sqrt{x}+3}/{\sqrt{x}-3}`
`=>4x=\sqrt{x}+3`
`=>4x-\sqrt{x}-3=0`
`=>4x-4\sqrt{x}+3\sqrt{x}-3=0`
`=>4\sqrt{x}(\sqrt{x}-1)+3(\sqrt{x}-1)=0`
`=>(\sqrt{x}-1)(4\sqrt{x}+3)=0`
`=>\sqrt{x}-1=0` (vì `4\sqrt{x}+3>0` với `x>0;x\ne 9)`
`=>\sqrt{x}=1`
`=>x=1` (thỏa mãn)
Vậy `x=1` thì `P={\sqrt{x}+3}/{\sqrt{x}-3}`
$\\$
`3)` `P={2x+4\sqrt{x}+2}/\sqrt{x}\quad (x>0;x\ne 1)`
Để `P=\sqrt{x}+7`
`=>{2x+4\sqrt{x}+2}/\sqrt{x}=\sqrt{x}+7`
`=>2x+4\sqrt{x}+2=(\sqrt{x}+7).\sqrt{x}`
`=>2x+4\sqrt{x}+2=x+7\sqrt{x}`
`=>x-3\sqrt{x}+2=0`
`=>x-2\sqrt{x}-\sqrt{x}+2=0`
`=>\sqrt{x}(\sqrt{x}-2)-(\sqrt{x}-2)=0`
`=>(\sqrt{x}-1)(\sqrt{x}-2)=0`
`=>`$\left[\begin{array}{l}\sqrt{x}=1\\\sqrt{x}=2\end{array}\right.$`=>`$\left[\begin{array}{l}x=1\ (loại)\\x=4\ (thỏa\ mãn)\end{array}\right.$
Vậy `x=4` thỏa mãn đề bài
$\\$
`4)` `A={x\sqrt{x}-1}/{x-1}\quad (x\ge 0;x\ne 1)`
`={(\sqrt{x})^3-1^3}/{x-1}`
`={(\sqrt{x}-1)(x+\sqrt{x}-1)}/{(\sqrt{x}-1)(\sqrt{x}+1)}`
`={x+\sqrt{x}+1}/{\sqrt{x}+1}`
$\\$
`B={x-4\sqrt{x}+4}/{x-4}\quad (x\ge 0;x\ne 4)`
`={(\sqrt{x}-2)^2}/{(\sqrt{x}-2)(\sqrt{x}+2)}`
`={\sqrt{x}-2}/{\sqrt{x}+2}`
$\\$
Để $A=B$
`<=>{x+\sqrt{x}+1}/{\sqrt{x}+1}={\sqrt{x}-2}/{\sqrt{x}+2}` `(x\ge 0;x\ne 1;4)`
`<=>{x+\sqrt{x}+1}/{\sqrt{x}+1}-{\sqrt{x}-2}/{\sqrt{x}+2}=0`
`<=>{(x+\sqrt{x}+1)(\sqrt{x}+2)-(\sqrt{x}-2)(\sqrt{x}+1)}/{(\sqrt{x}+1)(\sqrt{x}+2)}=0`
`<=>x\sqrt{x}+2x+x+2\sqrt{x}+\sqrt{x}+2-(x+\sqrt{x}-2\sqrt{x}-2)=0`
`<=>x\sqrt{x}+2x+4\sqrt{x}+4=0` $(1)$
Với mọi `x\ge 0;x\ne 1; x\ne 4` ta có:
`\qquad x\sqrt{x}+2x+4\sqrt{x}+4\ge 4>0`
`=>(1)` vô nghiệm
`=>` Không có giá trị của `x` để `A=B`
