Đáp án:
`a)` `A={x+3}/{\sqrt{x}+1}` với `x\ge 0; x\ne 4`
`b)` `A=4` khi `x=9+4\sqrt{5}`
`c)` `{\sqrt{5}+3}/{\sqrt{\sqrt{5}}+1}`
Giải thích các bước giải:
`a)` `A=({x+5\sqrt{x}-2}/{x-\sqrt{x}-2}+{x+1}/{\sqrt{x}+1}-{\sqrt{x}+2}/{\sqrt{x}-2})`
`\qquad (x\ge 0;x\ne 4)`
`A={x+5\sqrt{x}-2+(x+1)(\sqrt{x}-2)-(\sqrt{x}+2)(\sqrt{x}+1)}/{(\sqrt{x}+1)(\sqrt{x}-2)}`
`A={x+5\sqrt{x}-2+x\sqrt{x}-2x+\sqrt{x}-2-(x+\sqrt{x}+2\sqrt{x}+2)}/{(\sqrt{x}+1)(\sqrt{x}-2)}`
`A={-2x+x\sqrt{x}+3\sqrt{x}-6}/{(\sqrt{x}+1)(\sqrt{x}-2)}`
`A={x(\sqrt{x}-2)+3(\sqrt{x}-2)}/{(\sqrt{x}+1)(\sqrt{x}-2)}`
`A={(\sqrt{x}-2)(x+3)}/{(\sqrt{x}+1)(\sqrt{x}-2)}`
`A={x+3}/{\sqrt{x}+1}`
Vậy `A={x+3}/{\sqrt{x}+1}` với `x\ge 0; x\ne 4`
$\\$
`b)` Để `A=4`
`<=>{x+3}/{\sqrt{x}+1}=4` $(x\ge 0;x\ne 4)$
`<=>x+3=4(\sqrt{x}+1)`
`<=>x+3=4\sqrt{x}+4`
`<=>x-4\sqrt{x}=1`
`<=>x-4\sqrt{x}+4=5`
`<=>(\sqrt{x}-2)^2=5`
`<=>`$\left[\begin{array}{l}\sqrt{x}-2=\sqrt{5}\\\sqrt{x}-2=-\sqrt{5}\end{array}\right.$
`<=>`$\left[\begin{array}{l}\sqrt{x}=2+\sqrt{5}\\\sqrt{x}=2-\sqrt{5}<0\ (loại)\end{array}\right.$
`\qquad \sqrt{x}=2+\sqrt{5}`
`=>x=(2+\sqrt{5})^2=4+4\sqrt{5}+5`
`=>x=9+4\sqrt{5}\ (thỏa\ mãn)`
Vậy để `A=4` thì `x=9+4\sqrt{5}`
$\\$
`c)` $x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$
`=>`$x^3=(\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2})^3$
`=>`$x^3=\sqrt{5}+2+3.(\sqrt[3]{\sqrt{5}+2})^2 .\sqrt[3]{\sqrt{5}-2}+3\sqrt[3]{\sqrt{5}+2}.(\sqrt[3]{\sqrt{5}-2})^2+\sqrt{5}-2$
`=>`$x^3=2\sqrt{5}+3\sqrt[3]{(\sqrt{5}+2).(\sqrt{5}-2)}.(\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2})$
`=>`$x^3=2\sqrt{5}+3\sqrt[3]{5-2^2}.x$
`=>x^3=2\sqrt{5}+3.1.x`
`=>x^3-3x-2\sqrt{5}=0`
`=>x^3-5x+2x-2\sqrt{5}=0`
`=>x(x^2-5)+2(x-\sqrt{5})=0`
`=>x(x+\sqrt{5})(x-\sqrt{5})+2(x-\sqrt{5})=0`
`=>(x-\sqrt{5})(x^2+x\sqrt{5}+2)=0`
`=>(x-\sqrt{5})[(x+\sqrt{5}/2)^2+3/4]=0`
`=>x-\sqrt{5}=0` (vì `(x+\sqrt{5}/2)^2+3/4\ge 3/4>0` với mọi `x`)
`=>x=\sqrt{5}`
Với `x=\sqrt{5}\ (thỏa đk)`
`=> A={x+3}/{\sqrt{x}+1}={\sqrt{5}+3}/{\sqrt{\sqrt{5}}+1}`
Vậy: `A={\sqrt{5}+3}/{\sqrt{\sqrt{5}}+1}`
